How can one show $\displaystyle\frac{(m+n1)!}{m ! (n1)!}\leq \left[\frac{e (m+n1)}{n}\right]^{n1}$ ?

Denote the quotient of the right and left hand sides, $$ f(m,n)=\biggl(\frac{e(m+n1)}n\biggr)^{n1}\bigg/\binom{m+n1}m. $$ Then $f(m,1)=1$ for all $m\in\mathbb N$ and $$ \frac{f(m,n+1)}{f(m,n)} =\frac{e}{\biggl(1+\dfrac1n\biggr)^n}\cdot\biggl(1+\frac1{m+n1}\biggr)^{n1} > 1, $$ that is, $$ f(m,n+1)> f(m,n)>\dots> f(m,2)> f(m,1)=1. $$ This proves the required inequality. 


Another immediate proof can be obtained from $$ \frac{(m+n1)!}{m!}\le(m+n1)^{n1} $$ (which is obvious) and $$ \left(\frac{n}{e}\right)^{n1}\le(n1)! $$ which after multiplying by $n$ and taking logs becomes $$ n\log(n)n+1\le\sum_{k=2}^{n}\log(k) $$ which is immediate as the RHS is an obvious upper bound for $$ \int_1^n\log(x)\,dx=(x\log(x)x)_1^n=n\log(n)n+1. $$ 

