<s>[INST] Binomial Coefficients inequality simplifying to answer involving "e" In doing some analysis, my professor wrote \begin{align*} \begin{pmatrix}  n+ m \\ m \end{pmatrix}  &= \frac{(n+m)!}{m!n!} \\ &= \frac{(n+m)(n+m-1)...(n+1)n!} {m!n!} \\ &= \frac{(n+m)}{m} \frac{(n+m-1)}{m-1}  ... \frac{(n+1)}{1}  \\ & \leq \bigg( \frac{e(n+m)}{m} \bigg)^m \end{align*}
I understand the first equality is the definition of binomial coefficients, the third equality cancels the $n!$.  I am confused about how the last inequality.  I understand that $$ e^m = \sum_{k=0}^{\infty} \frac{m^k}{k!} $$ but I don't quite see how I get the last inequality from this.  If someone could tell me the explicit step I need to take I would greatly appreciate it.
 [/INST] The following inequality chain is valid for positive integers $m$
\begin{align*}
  \sqrt{2\pi m}\left(\frac{m}{e}\right)^m\leq m! \leq e\sqrt{m}\left(\frac{m}{e}\right)^m\tag{1}
  \end{align*}
The reciprocal of the left part of (1) gives
\begin{align*}
\frac{1}{m!}\leq\frac{1}{\sqrt{2\pi m}}\left(\frac{e}{m}\right)^m\leq\left(\frac{e}{m}\right)^m\tag{2}
\end{align*}
We also have
\begin{align*}
(n+m)(n+m-1)\cdots(n+1)\leq (n+m)^{m}\tag{3}
\end{align*}

From(2) and (3) we obtain
  \begin{align*}
\color{blue}{\frac{(n+m)(n+m-1)\cdots(n+1)}{m!}}&\leq \left(\frac{e}{m}\right)^m(n+m)^m\\
&\,\,\color{blue}{=\left(\frac{e(n+m)}{m}\right)^m}
\end{align*}
and the claim follows.

</s>