How to use Stirling's formula of n! in this probability computations in random walk? I want to compute $ \binom{2n}{n} p^n (1-p)^n = \frac{(2n)!}{n!n!}(p(1-p))^n, n=1,2,3...$
By using an approximation, due to Stirling, which asserts $ n! \sim n^{(n +\frac12)}e^{-n}\sqrt{2\pi}$
Where we say that $a_n \sim b_n$ when $ \lim_{(n\rightarrow \infty)} \frac{a_n}{b_n}=1,$ we obtain
$\frac{(2n)!}{n!n!} \sim \frac{(4p(1-p))^n}{\sqrt{n\pi}}\tag{1}$
I know $\frac{(2n)!}{n!}= 2^n \prod\limits_{n=0}^{n-1}(2n+1) = 2^n(1\cdot 3\cdot 5\cdot ...\cdot(2n -1))$
We can write $ \frac{(2n)!}{n!n!} = \frac{2^n( 1\cdot 3\cdot 5\cdot...\cdot(2n-1))}{n^n \sqrt{2} \sqrt{n\pi}e^{-n}}$
Now, how to further simplify it to get (1) when $n\rightarrow \infty$?
Now, how did author compute (1)?

I presume that you rather want to show:
$$
\frac{(2n)!}{n!n!} \sim \frac{4^n}{\sqrt{n\pi}}\tag{1}.
$$
Applying Stirling we have:
$$
\begin{align}
\frac{(2n)!}{n!n!} &\sim \frac{(2n)^{(2n + \frac12)}e^{-2n}\sqrt{2\pi}}{n^{2(n +\frac12)}e^{-2n}2\pi}\\
    &= \frac{2^{2n+\frac12}}{\sqrt{2\pi}} \frac{n^{2n+\frac12}}{n^{2n+1}}\\
    &= \frac{2^{2n}}{\sqrt\pi} n^{-\frac12}\\
    &= \frac{4^n}{\sqrt{\pi n}}.
\end{align}
$$
