factorial - Stirling's Formula normalization

An obvious way to get estimates on $n!$ is to compare $\sum\log k$ to $\int\log t$. If one could get Stirling's formula this way that would strike me as the "right" proof, because it would be clear why it works.

This morning I came much closer to this than I have in the past; in fact fairly straightforward comparisons of sums to integrals show that $$n!\sim c\sqrt n\left(\frac ne\right)^n.$$

Question: I wonder if there's some cheap trick to show that if $n!\sim c\sqrt n(n/e)^n$ then $c=\sqrt{2\pi}$.

Answer

You could calculate the normalization for the Gaussian approximation to the binomial distribution.