Entropy of a binomial distribution 
How do we get the functional form for the entropy of a binomial distribution? Do we use Stirling's approximation? 

According to Wikipedia, the entropy is:

$$\frac1 2 \log_2 \big( 2\pi e\, np(1-p) \big) + O \left( \frac{1}{n} \right)$$

As of now, my every attempt has been futile so I would be extremely appreciative if someone could guide me or provide some hints for the computation. 
 A: This answer follows roughly the suggestion of @MichaelLugo in the comments. 
We are interested in the sum 
$$H = -\sum_{k=0}^n {n\choose k}p^k(1-p)^{n-k}
\log_2\left[{n\choose k}p^k(1-p)^{n-k} \right].$$
For $n$ large we can use the de-Moivre-Laplace theorem, 
$$H \simeq 
-\int_{-\infty}^\infty dx \, 
\frac{1}{\sqrt{2\pi}\sigma} \exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right]
\log_2\left\{\frac{1}{\sqrt{2\pi}\sigma} \exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right] \right\},$$
where $\mu = n p$ and $\sigma^2 = n p(1-p)$.
Thus, 
$$\begin{eqnarray*}
H &\simeq& 
\int_{-\infty}^\infty dx \, 
\frac{1}{\sqrt{2\pi}\sigma} \exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right]
\left[\log_2(\sqrt{2\pi}\sigma) + \frac{(x-\mu)^2}{2\sigma^2} \log_2 e \right] \\
&=& \log_2(\sqrt{2\pi}\sigma) + \frac{\sigma^2}{2\sigma^2} \log_2 e  \\
&=& \frac{1}{2} \log_2 (2\pi e\sigma^2)
\end{eqnarray*}$$
and so 
$$H \simeq \frac{1}{2} \log_2 \left[2\pi e n p(1-p)\right].$$
Higher order terms can be found, essentially by deriving a more careful (and less simple) version of de-Moivre-Laplace.
