First I must show $${n \choose k} p^k q^{n-k} \sim \frac{1}{\sqrt{2\pi n p q}}e^{-\frac{(k-np)^2}{2npq}}?$$ How do I do this?
I know that Stirling's formula says that as $$n \rightarrow \infty.$$, $$n! \sim n^n e^{-n} \sqrt{2\pi n}.$$
I'm trying to prove the central limit theorem for binomial random variables and this is a step to do it from scratch.
To begin with: If $p$ happens to equal $\frac kn$, we readily get $$\begin{align}{n\choose k}p^kq^{n-k} &=\frac{n!}{k!(n-k)!}\cdot \frac{k^k}{n^k}\cdot \frac{(n-k)^{n-k}}{n^{n-k}}\\ &=\frac{\frac{n!}{n^n}}{\frac{k!}{k^k}\cdot \frac{(n-)k!}{(n-k)^{n-k}}}\\ &\approx \frac{e^{-n}\sqrt{2\pi n}}{e^{-k}\sqrt{2\pi k}\cdot e^{-(n-k)}\sqrt{2\pi(n-k)}}\\ &=\sqrt{\frac{n}{2\pi k(n-k)}}\\ &=\frac1{\sqrt{2\pi npq}}\end{align} $$ as desired.
Now see what happens when $p$ deviates from $\frac kn$. (it may be useful to work with the logarithm of the desired expression).