Can this be calculated by Stirling's formula?

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I'm reading Feller's book "An introduction to probability theory and its applications" chap 7.3, in this section he calculated the following binomial coefficient $$a_{k}=b(m+k ; n, p)=\left(\begin{array}{c} n \\ m+k \end{array}\right) p^{m+k} q^{n-m-k}$$ where $m=n p+\delta \ \text { with } -q<\delta \leq p$ and $p+q=1$. Finally he simplifies this into $$a_{k}=a_{0} e^{-k^{2} /(2 n p q)+\rho_{k}}$$ and $$a_{0}=\frac{n !}{m !(n-m) !} p^{m} q^{n-m} \sim \frac{1}{\sqrt{2 \pi n p q}}$$ where $\rho_{k}$ is error term and $\delta$ is neglected during the calculation.

My question is can this be simplified by using Stirling's formula? I was trying to calculate but I don't know where the term $k^{2}/2npq$ in the exponential comes from.