Binomial Coefficients inequality simplifying to answer involving "e"

Question

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.

Answer 1

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.