${n \choose k} \le ({en \over k})^k$ proof $${n \choose k} \le \left({en \over k}\right)^k$$
Could anyone give me a hint how to prove this by induction on $k$? (I can prove it without induction)
 A: General approach for proof by induction on $k$, for a given $n$:

$${n \choose k} \le \left({en \over k}\right)^k\tag{1}$$

First, we establish the base case $P(1)$: let $k = 1$ and show that the inequality holds.
Then we assume the inductive hypothesis, that is, assume the truth of $P(k)$, as given in $(1)$.
Then, using the inductive hypothesis, you test to see whether  $$P(k+1) = {n \choose k+1} \le \left({en \over k+1}\right)^{k+1}$$ 
The challenging part is the last: use what you know (having proved it without induction) to write $P(k+1)$ in terms of the assumed inductive hypothesis. Then it's simply manipulating to exhibit the final form of the inequality you are seeking to establish.

Update: You're on your way, and almost there, with your inductive step. Recall, as robjohn notes below your post, you want to prove $$\binom{n}{k+1} \leq \left(\frac{en}{k+1}\right)^{k+1}.$$
Making a few tweaks with respect to your work, we have that:
$$
\begin{align} \\
\binom{n}{k+1}
& = \frac{n}{k+1}\binom{n-1}{k}  \\ \\
& \leq \frac{n}{k+1} \left(\frac{e(n-1)}{k}\right)^k \quad\quad\quad\tag{inductive hypothesis} \\ \\
& = \frac{n}{k+1}\left(\frac{k+1}{k}\right)^k \left(\frac{e(n - 1)}{k+1}\right)^k \\ \\
& \leq \frac{en}{k+1} \left(\frac{e(n-1)}{k+1}\right)^k \\ \\
& \leq \frac{en}{k+1} \left(\frac{en}{k+1}\right)^k \\ \\
& = \left(\frac{en}{k+1}\right)^{k+1} \\ \\
\end{align}$$
A: Induction step:
$${n \choose k+1}={n\over k+1}{n-1 \choose k} \le {n\over k+1} \left({e(n-1) \over k}\right)^k \le {n\over k} \left({en \over k}\right)^k \le \left({en \over k}\right)^{k+1}$$
A: Looking back at the answers that have been posted while I was working on this, I see that this is similar to AmWhy's argument. I will post it anyway since there is more detail (and color).
The base case is
$$
\binom{n}{1}=n\le en=\left(\frac{en}{1}\right)^1\tag{1}
$$
We avoid $k=0$ due to division by $0$.
Suppose the inequality is true for $\displaystyle\binom{n}{k}$, then
$$
\begin{align}
\binom{n+1}{k+1}
&=\frac{n+1}{k+1}\binom{n}{k}\\
&\color{#C00000}{\le}\frac{n+1}{k+1}\color{#C00000}{\left(\frac{en}{k}\right)^k}\\
&=\frac{n+1}{k+1}\left(\frac{k+1}{k}\right)^k\left(\frac{en}{k+1}\right)^k\\
&\color{#C00000}{\le}\frac{n+1}{k+1}\color{#C00000}{e}\left(\frac{en}{k+1}\right)^k\\
&\color{#C00000}{\le}\frac{n+1}{k+1}e\left(\frac{e\color{#C00000}{(n+1)}}{k+1}\right)^k\\
&=\left(\frac{e(n+1)}{k+1}\right)^{k+1}\tag{2}
\end{align}
$$
Using $(1)$ and $(2)$, we see that
$$
\binom{n}{k}\le\left(\frac{en}{k}\right)^k\tag{3}
$$
For $n\ge k\ge1$.
