How prove $\binom{n}{m}\le\left(\frac{en}{m}\right)^m$

Question

Show that $$\binom{n}{m}\le\left(\dfrac{en}{m}\right)^m$$ where $0<m\le n,m,n\in N^{+}$

My idea: since $$(\dfrac{n}{m}-\dfrac{m-1}{m})(\dfrac{n}{m}-\dfrac{m-2}{m})\cdots\dfrac{n}{m}\le\left(\dfrac{en}{m}\right)^m$$ $$\sum_{i=1}^{m}\ln{\left(\dfrac{n}{m}-\dfrac{m-i}{m}\right)}\le m\ln{\dfrac{en}{m}}$$

then I can't,Thank you

Answer 1

First note that

$${n\choose m}=\frac1{m!}\prod_{k=n-m+1}^nk\le\frac1{m!}\prod_{k=n-m+1}^nn=\frac{n^m}{m!}.$$

Furthermore, the bound form of Stirling's approximation is

$$m!\ge\sqrt{2\pi m}\left(\frac me\right)^m\ge\left(\frac me\right)^m,$$

so

$${n\choose m}\le\frac{n^m}{m!}\le\left(\frac{en}m\right)^m.$$