Inequality involving $e$: $\binom{n}{k} < \frac{1}{e}\big(\frac{en}{k}\big)^k$

Question

$\forall n,k\in \mathbb{Z^+}$, prove that $\binom{n}{k} < \frac{1}{e}\big(\frac{en}{k}\big)^k$ where $e$ is the [Euler's number](https://en.wikipedia.org/wiki/E_(mathematical_constant). I tried my best to solve it and thought of expanding both sides but it helps me nowhere. Also, the $k$th power seems to be clumsy, i cannot reach the conclusion. Please help me!

Answer 1

You have: $$\binom{n}{k}=\frac{n!}{k!(n-k)!} = \frac{n(n-1)\cdots(n-k+1)}{k!}\leq \frac{n^k}{k!}.$$ Now we need a lower bound for $k!$. We will use $k! \geq \frac{k^k}{e^{k-1}}$, which you can see the proof of here. Using this, $$\binom{n}{k} \leq \frac{n^k}{k!} \leq \frac{n^k\cdot e^{k-1}}{k^k}= \frac{1}{e}\left(\frac{en}{k}\right)^k.$$