TITLE: Stirling's Approximation for binomial coefficient
QUESTION [19 upvotes]: In this proof, it is assumed that, for $k \ll n$, ${n \choose k} \approx \frac{n^k}{k!}$, given Stirling's approximation.
How does Stirling's Approximation, in either form $\ln n! \approx n\ln{n} -n + (\ln(n))$ or $n! \approx \sqrt{2\pi n} (\frac{n}{e})^n$, give this result?

REPLY [7 votes]: The approximation $n! \approx (n/e)^n$ suffices. As $n \to \infty$ and $k/n \to 0$ we have 
$$(n-k)! \approx \left(\frac{n-k}{e}\right)^{n-k}= \left(\frac{n}{e}\right)^{n-k} (1-k/n)^{n-k} \approx\left(\frac{n}{e}\right)^{n-k} e^{-k}$$
Hence 
$$ 
{n \choose k} \approx \frac{ \left(\frac{n}{e}\right)^{n}} {\left(\frac{n}{e}\right)^{n-k} e^{-k} \, k!} = \frac{n^k}{k!}
$$