We want to show that for $1\leq k\leq n$, the following holds: ${n\choose k} \leq \Big(\frac{n\cdot e}{k}\Big)^k.$

First, recall the power series $e^k=\sum_{i=0}^\infty{\frac{k^i}{i!}}$. For $k \geq 1$, this sum is definitely larger than its $k$th term only. In other words, $e^k \geq k^k/k!$, which implies $k! \geq (k/e)^k$. Now,

$\begin{matrix}{n\choose k} &=& \frac{n!}{k! \cdot (n-k)!}\\ &=& \frac{n(n-1)(n-2)\cdots (n-(k-1))}{k!}\\ &\leq& \frac{n^k}{k!} \\ &\leq& \frac{n^k}{(k/e)^k} \\ &=& (\frac{en}{k})^k.\\ \end{matrix}$

This proof is extracted from the StackExchange discussion here. For more inequalities/bounds for binomial coefficients, see Wikipedia.