Inequality ${n \choose k} \leq \left(\frac{en}{ k}\right)^k$ This is from page 3 of http://www.math.ucsd.edu/~phorn/math261/9_26_notes.pdf (Wayback Machine).
Copying the relevant segment:

Stirling’s approximation tells us $\sqrt{2\pi n} (n/e)^n \leq n! \leq e^{1/12n} \sqrt{2\pi n} (n/e)^n$. In particular we can use this to say that
  $$ {n \choose k} \leq \left(\frac{en}{ k}\right)^k$$

I tried the tactic of combining bounds from $n!$, $k!$ and $(n-k)!$ and it didn't work. How does this bound follow from stirling's approximation?

First of all, note that $n!/(n-k)! \le n^k$. Use Stirling only for $k!$.
${n \choose k} \le \frac{n^k}{k!} \le \frac{n^k}{(\sqrt{2\pi k}(k/e)^k)} \le \frac{n^k}{(k/e)^k} = (\frac{en}{k})^k$ 
