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\begin{document}
\section*{Math 222 Fall 2022, Lecture 6: Introduction}

\textbf{website:}
\texttt{\href{https://www.cip.ifi.lmu.de/~grinberg/t/22fco}{\texttt{https://www.cip.ifi.lmu.de/\symbol{126}%
grinberg/t/22fco}}}

\section{Introduction (cont'd)}

\setcounter{subsection}{2}

\subsection{Factorials and binomial coefficients (cont'd)}

\setcounter{subsubsection}{2}

\subsubsection{Fundamental properties of the binomial coefficients (cont'd)}

\setcounter{theo}{4}

Last time, we proved a first observation about BCs (= binomial coefficients):

\begin{proposition}
\label{prop.binom.0}If $n\in\mathbb{N}$ and $k\in\mathbb{R}$ satisfy $k>n$,
then $\dbinom{n}{k}=0$.
\end{proposition}

Here are some others:\footnote{In the following, I will only sketch the
proofs. Detailed proofs can always be found in the 2019 notes (\S 1.3 to be
specific).}

\begin{proposition}
[Upper negation formula]\label{prop.binom.upneg}Let $n\in\mathbb{R}$ and
$k\in\mathbb{Z}$. Then,%
\[
\dbinom{-n}{k}=\left(  -1\right)  ^{k}\dbinom{n+k-1}{k}.
\]

\end{proposition}

\begin{proof}
If $k\notin\mathbb{N}$, then this boils down to $0=\left(  -1\right)
^{k}\cdot0$, which is clear.

So WLOG assume that $k\in\mathbb{N}$. Then, the definition of binomial
coefficients yields%
\begin{align*}
\dbinom{-n}{k}  &  =\dfrac{\left(  -n\right)  \left(  -n-1\right)  \left(
-n-2\right)  \cdots\left(  -n-k+1\right)  }{k!}\ \ \ \ \ \ \ \ \ \ \text{and}%
\\
\dbinom{n+k-1}{k}  &  =\dfrac{\left(  n+k-1\right)  \left(  n+k-2\right)
\left(  n+k-3\right)  \cdots n}{k!}.
\end{align*}
These two fractions are equal up to a $\left(  -1\right)  ^{k}$ factor, because

\begin{itemize}
\item their denominators are the same, and

\item their numerators are the same product, except with its factors reversed
and with each factor negated (which is how we get the $\left(  -1\right)
^{k}$).
\end{itemize}

Thus, the upper negation formula (Proposition \ref{prop.binom.upneg}) is proved.
\end{proof}

The upper negation formula can be used to reduce questions about binomial
coefficients $\dbinom{n}{k}$ with $n\in\left\{  -1,-2,-3,\ldots\right\}  $ to
questions about those with $n\in\mathbb{N}$. Thus, most tables of binomial
coefficients that appear in the literature omit the part with $n<0$ and focus
on the part with $n\geq0$. This latter part is known as \textbf{Pascal's
triangle}. The zeroes to the right of the $k=n$ line (these are the zeroes
guaranteed by Proposition \ref{prop.binom.0}) are also omitted. Here is a
table of Pascal's triangle with a few more rows than we saw last time:%
\[
\scalebox{1}{
\setlength{\tabcolsep}{1.3pt}
\begin{tabular}
[c]{ll|ccccccccccccccccc}\phantom{$\overset{k=0}{\swarrow}$} &  &
\phantom{$\overset{k=0}{\swarrow}$} &
\phantom{$\overset{k=0}{\swarrow}$} &
\phantom{$\overset{k=0}{\swarrow}$} &
\phantom{$\overset{k=0}{\swarrow}$} &
\phantom{$\overset{k=0}{\swarrow}$} &
\phantom{$\overset{k=0}{\swarrow}$} &
\phantom{$\overset{k=0}{\swarrow}$} &
\phantom{$\overset{k=0}{\swarrow}$} &
\phantom{$\overset{k=0}{\swarrow}$} &
$\overset{k=0}{\swarrow}$ & $\phantom{20}$ & $\phantom{20}$ & $\phantom{20}$ &
$\phantom{20}$ & $\phantom{20}$ & $\phantom{20}$ & $\phantom{20}$\\
\vphantom{$\overset{k=0}{\swarrow}$}$n=0$ & $\rightarrow$ &  &  &  &  &  &  &
&  & $1$ &  & $\overset{k=1}{\swarrow}$ &  &  &  &  &  & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=1$ & $\rightarrow$ &  &  &  &  &  &  &
& $1$ &  & $1$ &  & $\overset{k=2}{\swarrow}$ &  &  &  &  & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=2$ & $\rightarrow$ &  &  &  &  &  &  &
$1$ &  & $2$ &  & $1$ &  & $\overset{k=3}{\swarrow}$ &  &  &  & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=3$ & $\rightarrow$ &  &  &  &  &  & $1$
&  & $3$ &  & $3$ &  & $1$ &  & $\overset{k=4}{\swarrow}$ &  &  & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=4$ & $\rightarrow$ &  &  &  &  & $1$ &
& $4$ &  & $6$ &  & $4$ &  & $1$ &  & $\overset{k=5}{\swarrow}$ &  & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=5$ & $\rightarrow$ &  &  &  & $1$ &  &
$5$ &  & $10$ &  & $10$ &  & $5$ &  & $1$ &  & $\overset{k=6}{\swarrow}$ & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=6$ & $\rightarrow$ &  &  & $1$ &  & $6$
&  & $15$ &  & $20$ &  & $15$ &  & $6$ &  & $1$ &  & $\overset{k=7}{\swarrow}$\\
\vphantom{$\overset{k=0}{\swarrow}$}$n=7$ & $\rightarrow$ &  & $1$ &  & $7$ &
& $21$ &  & $35$ &  & $35$ &  & $21$ &  & $7$ &  & $1$ & \\
\vphantom{$\overset{k=0}{\swarrow}$}$n=8$ & $\rightarrow$ & $1$ &  & $8$ &  &
$28$ &  & $56$ &  & $70$ &  & $56$ &  & $28$ &  & $8$ &  & $1$\end{tabular}
}\ \ \ \ \ \
\]


\bigskip

Now here are some more properties of Pascal's triangle. The following theorem
(known as \textbf{Pascal's recurrence}, or as the \textbf{recurrence of the
binomial coefficients}) shows that each entry of this triangle is the sum of
its two neighbor entries above it:

\begin{theorem}
[Pascal's recurrence]\label{thm.binom.rec}For any $n\in\mathbb{R}$ and
$k\in\mathbb{R}$, we have%
\[
\dbinom{n}{k}=\dbinom{n-1}{k-1}+\dbinom{n-1}{k}.
\]

\end{theorem}

\begin{proof}
Three cases are possible:

\textit{Case 1:} We have $k\notin\mathbb{N}$.

\textit{Case 2:} We have $k=0$.

\textit{Case 3:} We have $k\in\left\{  1,2,3,\ldots\right\}  $.

In Case 1, the claim boils down to $0=0+0$.

In Case 2, the claim boils down to $1=0+1$.

In Case 3, both $k$ and $k-1$ belong to $\mathbb{N}$. Thus, the definition of
BCs (= binomial coefficients) yields%
\begin{align*}
\dbinom{n}{k}  &  =\dfrac{n\left(  n-1\right)  \left(  n-2\right)
\cdots\left(  n-k+1\right)  }{k!};\\
\dbinom{n-1}{k-1}  &  =\dfrac{\left(  n-1\right)  \left(  n-2\right)  \left(
n-3\right)  \cdots\left(  n-k+1\right)  }{\left(  k-1\right)  !};\\
\dbinom{n-1}{k}  &  =\dfrac{\left(  n-1\right)  \left(  n-2\right)  \left(
n-3\right)  \cdots\left(  n-k\right)  }{k!}.
\end{align*}
Bringing these fractions to a common denominator, we obtain%
\begin{align*}
\dbinom{n}{k}  &  =\dfrac{n\left(  n-1\right)  \left(  n-2\right)
\cdots\left(  n-k+1\right)  }{k!};\\
\dbinom{n-1}{k-1}  &  =\dfrac{k\cdot\left(  n-1\right)  \left(  n-2\right)
\left(  n-3\right)  \cdots\left(  n-k+1\right)  }{k!};\\
\dbinom{n-1}{k}  &  =\dfrac{\left(  n-1\right)  \left(  n-2\right)  \left(
n-3\right)  \cdots\left(  n-k\right)  }{k!}%
\end{align*}
(since $k!=\left(  k-1\right)  !\cdot k$, as we saw last time). So it remains
to prove that%
\begin{align*}
&  n\left(  n-1\right)  \left(  n-2\right)  \cdots\left(  n-k+1\right) \\
&  =k\cdot\left(  n-1\right)  \left(  n-2\right)  \left(  n-3\right)
\cdots\left(  n-k+1\right)  +\left(  n-1\right)  \left(  n-2\right)  \left(
n-3\right)  \cdots\left(  n-k\right)  .
\end{align*}
After cancelling common factors, this boils down to%
\[
n=k+\left(  n-k\right)  ,
\]
which is true. So the recurrence formula (Theorem \ref{thm.binom.rec}) is proved.
\end{proof}

Next comes a very short formula for binomial coefficients, which however works
only under specific conditions:

\begin{theorem}
[factorial formula for BCs]\label{thm.binom.fac}Let $n\in\mathbb{N}$ and
$k\in\mathbb{N}$ be such that $k\leq n$. Then,%
\[
\dbinom{n}{k}=\dfrac{n!}{k!\cdot\left(  n-k\right)  !}.
\]

\end{theorem}

\begin{proof}
The definition of BCs yields%
\begin{align*}
&  k!\cdot\left(  n-k\right)  !\cdot\dbinom{n}{k}\\
&  =k!\cdot\left(  n-k\right)  !\cdot\dfrac{n\left(  n-1\right)  \left(
n-2\right)  \cdots\left(  n-k+1\right)  }{k!}\\
&  =\left(  \left(  n-k\right)  !\right)  \cdot\left(  n\left(  n-1\right)
\left(  n-2\right)  \cdots\left(  n-k+1\right)  \right) \\
&  =\left(  1\cdot2\cdot\cdots\cdot\left(  n-k\right)  \right)  \cdot\left(
n\left(  n-1\right)  \left(  n-2\right)  \cdots\left(  n-k+1\right)  \right)
\\
&  =1\cdot2\cdot\cdots\cdot n\ \ \ \ \ \ \ \ \ \ \left(  \text{by reordering
the factors of the product}\right) \\
&  =n!,
\end{align*}
qed.
\end{proof}

\textbf{Warning:} The factorial formula $\dbinom{n}{k}=\dfrac{n!}%
{k!\cdot\left(  n-k\right)  !}$ only works when $n\in\mathbb{N}$ and
$k\in\mathbb{N}$ and $k\leq n$. It does not help you compute $\dbinom{-1}{1}$
or $\dbinom{\sqrt{2}}{2}$ or $\dbinom{2}{3}$. It is nice when it works, but it
has its limits!

\bigskip

Here is another property of BCs, which manifests itself on Pascal's triangle
as a vertical symmetry (across the vertical axis in the middle of the triangle):

\begin{theorem}
[symmetry of binomial coefficients]\label{thm.binom.symm}For any
$n\in\mathbb{N}$ and $k\in\mathbb{R}$, we have%
\[
\dbinom{n}{k}=\dbinom{n}{n-k}.
\]

\end{theorem}

\begin{proof}
Let $n\in\mathbb{N}$ and $k\in\mathbb{R}$. We are in one of the following cases:

\textit{Case 1:} We have $k\notin\mathbb{N}$.

\textit{Case 2:} We have $k\in\mathbb{N}$ but $k>n$.

\textit{Case 3:} We have $k\in\mathbb{N}$ and $k\leq n$.

In Case 1, the claim boils down to $0=0$ (why?).

In Case 2, the claim boils down to $0=0$ (why?).

In Case 3, we can apply the factorial formula and obtain%
\begin{align*}
\dbinom{n}{k}  &  =\dfrac{n!}{k!\cdot\left(  n-k\right)  !}%
\ \ \ \ \ \ \ \ \ \ \text{and}\\
\dbinom{n}{n-k}  &  =\dfrac{n!}{\left(  n-k\right)  !\cdot\left(  n-\left(
n-k\right)  \right)  !}=\dfrac{n!}{\left(  n-k\right)  !\cdot k!}=\dfrac
{n!}{k!\cdot\left(  n-k\right)  !},
\end{align*}
which is of course the same number.
\end{proof}

Again, a \textbf{warning:} The symmetry $\dbinom{n}{k}=\dbinom{n}{n-k}$ does
not hold for negative $n$ or for non-integer $n$. For example, check that it
fails for $n=-1$ and $k=0$.

\bigskip

\subsubsection{Binomial coefficients count subsets}

The combinatorial meaning of BCs (binomial coefficients) is provided by the
following theorem:

\begin{theorem}
[combinatorial interpretation of BCs]\label{thm.binom.comb}Let $n\in
\mathbb{N}$ and $k\in\mathbb{R}$. Let $S$ be an $n$-element set. Then,%
\[
\dbinom{n}{k}=\left(  \text{\# of }k\text{-element subsets of }S\right)  .
\]

\end{theorem}

\begin{example}
\textbf{(a)} Let $n=4$ and $k=2$ and $S=\left\{  1,2,3,4\right\}  $. Then,
Theorem \ref{thm.binom.comb} says that
\[
\dbinom{4}{2}=\left(  \text{\# of }2\text{-element subsets of }S\right)  .
\]
Let's check this: We have $\dbinom{4}{2}=6$, and the $2$-element subsets of
$S$ are%
\[
\left\{  1,2\right\}  ,\ \ \left\{  3,4\right\}  ,\ \ \left\{  1,4\right\}
,\ \ \left\{  1,3\right\}  ,\ \ \left\{  2,3\right\}  ,\ \ \left\{
2,4\right\}  .
\]


\textbf{(b)} Let $n$ and $S$ be as before, and let $k=5$. Then, Theorem
\ref{thm.binom.comb} says that%
\[
\dbinom{4}{5}=\left(  \text{\# of }5\text{-element subsets of }S\right)  .
\]
Let's check this: We have $\dbinom{4}{5}=0$, and $S$ has no $5$-element subsets.
\end{example}

\textbf{Warning:} Theorem \ref{thm.binom.comb} could be used as an alternative
definition of BCs, if we only cared about the case of $\dbinom{n}{k}$ for
$n\in\mathbb{N}$ exclusively. But it says nothing about $\dbinom{-1}{1}$ or
$\dbinom{\sqrt{2}}{2}$.

We shall prove Theorem \ref{thm.binom.comb} by induction on $n$. The base case
($n=0$) relies on computing $\dbinom{0}{k}$:

\begin{lemma}
\label{lem.binom.0k}Let $k\in\mathbb{R}$. Then, $\dbinom{0}{k}=\left[
k=0\right]  $.
\end{lemma}

Here we are using the so-called \textbf{Iverson bracket notation}:

\begin{definition}
\label{def.iverson}If $\mathcal{A}$ is any logical statement, then the integer
$\left[  \mathcal{A}\right]  $ is defined to be%
\[%
\begin{cases}
1, & \text{if }\mathcal{A}\text{ is true};\\
0, & \text{if }\mathcal{A}\text{ is false.}%
\end{cases}
\]
This integer $\left[  \mathcal{A}\right]  $ is called the \textbf{truth value}
of $\mathcal{A}$. For example, $\left[  1+1=2\right]  =1$ but $\left[
1+1=1\right]  =0$.
\end{definition}

\begin{proof}
[Proof of Lemma \ref{lem.binom.0k}.]Straightforward and LTTR (see Lemma 1.3.14
in the 2019 notes).
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.binom.comb}.]Induct on $n$:

\textit{Base case:} We need to prove Theorem \ref{thm.binom.comb} for $n=0$.
So let $n=0$, and let $S$ be a $0$-element set (i.e., the empty set). Then, we
must show that%
\[
\dbinom{0}{k}=\left(  \text{\# of }k\text{-element subsets of }S\right)  .
\]
But this is clear by computing both sides: The LHS (= left-hand side) is
$\left[  k=0\right]  $ (by Lemma \ref{lem.binom.0k}); the RHS (= right-hand
side) is also $\left[  k=0\right]  $, since the empty set $S$ has only one
subset, namely itself (which is a $0$-element subset). So Theorem
\ref{thm.binom.comb} is proved for $n=0$.

\textit{Step:} Let $m\in\mathbb{N}$. Assume as the IH (= induction hypothesis)
that Theorem \ref{thm.binom.comb} holds for $n=m$. We must prove it for
$n=m+1$.

So let $k\in\mathbb{R}$, and let $S$ be an $\left(  m+1\right)  $-element set.
We must prove that%
\[
\dbinom{m+1}{k}\overset{?}{=}\left(  \text{\# of }k\text{-element subsets of
}S\right)  .
\]
(The question mark above the equality sign here just signals that this
equality has not been proved yet.)

Since $S$ is nonempty (because $\left\vert S\right\vert =m+1>m\geq0$), we can
pick an element $t\in S$ and consider the $m$-element set $S\setminus\left\{
t\right\}  $. Let us do this. The IH can then be applied to $S\setminus
\left\{  t\right\}  $ instead of $S$, and we obtain%
\begin{align*}
\dbinom{m}{k}  &  =\left(  \text{\# of }k\text{-element subsets of }%
S\setminus\left\{  t\right\}  \right)  \ \ \ \ \ \ \ \ \ \ \text{and}\\
\dbinom{m}{k-1}  &  =\left(  \text{\# of }\left(  k-1\right)  \text{-element
subsets of }S\setminus\left\{  t\right\}  \right)  .
\end{align*}
(Note that the IH is a \textquotedblleft for all $k\in\mathbb{R}%
$\textquotedblright\ statement, so we can indeed apply it to $k-1$ instead of
$k$, which is precisely what we have done here to get the second of these two equalities.)

By Pascal's recurrence (Theorem \ref{thm.binom.rec}), we have%
\begin{align}
\dbinom{m+1}{k}  &  =\dbinom{m}{k-1}+\dbinom{m}{k}\nonumber\\
&  =\left(  \text{\# of }\left(  k-1\right)  \text{-element subsets of
}S\setminus\left\{  t\right\}  \right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \text{\# of }k\text{-element subsets of
}S\setminus\left\{  t\right\}  \right)  . \label{pf.thm.binom.comb.5}%
\end{align}
So we only need to show that this sum equals $\left(  \text{\# of
}k\text{-element subsets of }S\right)  $. At last, we're getting to do some combinatorics!

Here is how to show this: We call a subset of $S$

\begin{itemize}
\item \textbf{red} if it contains $t$, and

\item \textbf{green} if it does not contain $t$.
\end{itemize}

Thus, each subset of $S$ is either red or green, but not both. So by the sum
rule, we have%
\begin{align*}
&  \left(  \text{\# of }k\text{-element subsets of }S\right) \\
&  =\left(  \text{\# of red }k\text{-element subsets of }S\right)  +\left(
\text{\# of green }k\text{-element subsets of }S\right)  .
\end{align*}
Now we need to compute the two addends on the RHS.

The green subsets of $S$ are precisely the subsets of $S\setminus\left\{
t\right\}  $. So%
\begin{align*}
&  \left(  \text{\# of green }k\text{-element subsets of }S\right) \\
&  =\left(  \text{\# of }k\text{-element subsets of }S\setminus\left\{
t\right\}  \right)  .
\end{align*}


What about the red ones? They are not subsets of $S\setminus\left\{
t\right\}  $, since they do contain $t$. However, they are \textquotedblleft
essentially\textquotedblright\ $\left(  k-1\right)  $-element subsets of
$S\setminus\left\{  t\right\}  $, because if you remove the element $t$ from
them, you're left with $\left(  k-1\right)  $-element subsets of
$S\setminus\left\{  t\right\}  $. Let us make this rigorous: There is a
bijection (= one-to-one-correspondence)%
\begin{align*}
\left\{  \text{red }k\text{-element subsets of }S\right\}   &  \rightarrow
\left\{  \left(  k-1\right)  \text{-element subsets of }S\setminus\left\{
t\right\}  \right\}  ,\\
R  &  \mapsto R\setminus\left\{  t\right\}
\end{align*}
(its inverse map sends each $Q$ to $Q\cup\left\{  t\right\}  $). Informally,
this is just saying that picking a red $k$-element subset of $S$ is the same
as picking a $\left(  k-1\right)  $-element subset of $S\setminus\left\{
t\right\}  $ and inserting $t$ into it. By the bijection principle, this
bijection entails that%
\begin{align*}
&  \left(  \text{\# of red }k\text{-element subsets of }S\right) \\
&  =\left(  \text{\# of }\left(  k-1\right)  \text{-element subsets of
}S\setminus\left\{  t\right\}  \right)  .
\end{align*}


Combining what we have proved, we get%
\begin{align*}
&  \left(  \text{\# of }k\text{-element subsets of }S\right)  \\
&  =\left(  \text{\# of red }k\text{-element subsets of }S\right)  +\left(
\text{\# of green }k\text{-element subsets of }S\right)  \\
&  =\left(  \text{\# of }\left(  k-1\right)  \text{-element subsets of
}S\setminus\left\{  t\right\}  \right)  +\left(  \text{\# of }k\text{-element
subsets of }S\setminus\left\{  t\right\}  \right)  \\
&  =\dbinom{m+1}{k}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.thm.binom.comb.5})}\right)  .
\end{align*}
This is precisely what we wanted to show in order to complete the induction
step. Thus, Theorem \ref{thm.binom.comb} is proved by induction.
\end{proof}

\begin{fineprint}
We take a moment to mention some terminology that we will not use: When $S$ is
a set and $k$ is an integer, the $k$-element subsets of $S$ are sometimes
called the $k$\textbf{-combinations of }$S$ (or, more precisely, the
$k$\textbf{-combinations of }$S$ \textbf{without repetition}\footnote{The
\textquotedblleft$k$-combinations with repetition\textquotedblright\ are the
size-$k$ multisubsets of $S$, which we will define and study in \S 2.9.}).
This notation was widespread until the 1950s, when it fell out of use because
\textquotedblleft$k$-element subsets\textquotedblright\ (or \textquotedblleft%
$k$-subsets\textquotedblright\ for short) is clearer and more consistent with
other parts of mathematics. Nevertheless, be prepared to occasionally see the
old \textquotedblleft$k$-combinations\textquotedblright\ terminology in the literature.
\end{fineprint}

\subsubsection{Integrality and arithmetic properties}

Here is something that looks completely natural when you look at Pascal's
triangle, but actually is not obvious:

\begin{theorem}
[integrality of binomial coefficients]\label{thm.binom.int}For any
$n\in\mathbb{Z}$ and $k\in\mathbb{R}$, the number $\dbinom{n}{k}$ is an integer.
\end{theorem}

\begin{proof}
If $n\geq0$, then this number $\dbinom{n}{k}$ counts the $k$-element subsets
of $\left[  n\right]  $ (by Theorem \ref{thm.binom.comb}), and thus is
obviously an integer. The case $n<0$ can easily be reduced to the case
$n\geq0$ using upper negation. (See the 2019 notes for details, as well as for
an alternative proof.)
\end{proof}

\begin{proposition}
\label{prop.binom.pdiv}Let $p$ be a prime number, and let $k\in\left\{
1,2,\ldots,p-1\right\}  $. Then, $p\mid\dbinom{p}{k}$.
\end{proposition}

\begin{proof}
We need to know one basic fact about prime numbers: If the prime number $p$
divides a product $a_{1}a_{2}\cdots a_{m}$ of $m$ integers $a_{1},a_{2}%
,\ldots,a_{m}$, then it must divide (at least) one of these $m$ integers
$a_{1},a_{2},\ldots,a_{m}$. We shall call this fact \textquotedblleft%
\textbf{Euclid's lemma}\textquotedblright, although this name is more commonly
used for \href{https://en.wikipedia.org/wiki/Euclids_lemma}{the particular
case $m=2$} (but that doesn't make a big difference, since the general case
can easily be derived by induction from the $m=2$ case).

Now, $p$ divides none of the $k$ integers $1,2,\ldots,k$ (since all these $k$
integers are $>0$ but $<p$). Hence, $p$ cannot divide their product
$1\cdot2\cdot\cdots\cdot k$ either (as otherwise, Euclid's lemma would yield
that $p$ divides one of these $k$ integers). In other words, $p$ cannot divide
$k!$.

However, the definition of BCs yields
\[
\dbinom{p}{k}=\dfrac{p\left(  p-1\right)  \left(  p-2\right)  \cdots\left(
p-k+1\right)  }{k!},
\]
so that $k!\cdot\dbinom{p}{k}=p\left(  p-1\right)  \left(  p-2\right)
\cdots\left(  p-k+1\right)  $. This entails that $p$ divides the product
$k!\cdot\dbinom{p}{k}$ (since there is a $p$ factor on the RHS of this
equality). Hence, by Euclid's lemma, $p$ must divide at least one of the two
factors $k!$ and $\dbinom{p}{k}$. Since we already know that $p$ cannot divide
$k!$, we thus conclude that $p$ divides $\dbinom{p}{k}$. This proves
Proposition \ref{prop.binom.pdiv}.
\end{proof}


\end{document}