In Rick Durrett's book, in a proof for the asymptotic behaviour of Poisson to normal, he uses the following identity:
$$\frac{n!n^m}{(n+m)!} = \left(\prod_{k=1}^m 1 + k/n \right)^{-1}$$
I'm just wondering how this established? As Durrett states it without proof, and I am not sure how to show this.
For $n,m\in\Bbb N_0$, $m>n$ we define $$\begin{align} P(n,m)=&\prod_{k=1}^m\frac1{1+k/n}\\ =&\prod_{k=1}^m\frac{n}{n+k}\\ =&n^m\prod_{k=1}^m\frac{1}{n+k}\\ \end{align}$$ Then note that $$\begin{align} \prod_{k=1}^{m}\frac1{n+k}=&\frac{n!}{n!}\prod_{k=1}^{m}\frac1{n+k}\\ =&\frac{n!}{n!\prod_{k=1}^{m}(n+k)}\\ =&\frac{n!}{(\prod_{r=1}^{n}r)(\prod_{k=n+1}^{n+m}k)}\\ =&\frac{n!}{\prod_{r=1}^{n+m}r}\\ =&\frac{n!}{(n+m)!}\\ \end{align}$$ So $$P(n,m)=\frac{n!n^m}{(n+m)!}$$