Using Wallis' product to derive $\sqrt\pi$ Recall Wallis' product:
$$\lim_{n\to\infty}\Big(\frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\frac{6}{5}\cdots\frac{2n}{2n-1}\cdot\frac{2n}{2n+1}\Big)=\frac{\pi}{2}$$
We have to show that $$\lim_{n\to\infty}\frac{(n!)^22^{2n}}{(2n)!\sqrt n}=\sqrt\pi$$
The hint I got was to use $$P_n=\frac{(n!)^42^{4n}}{[(2n)!]^2(2n+1)}$$
which is just simply the inside of the limit in Wallis' product, multiplied and divided by $2\cdot2\cdot4\cdot4\cdots(2n)\cdot(2n)$ alternatively. How do I use $P_n$ to derive $\sqrt\pi\:$? 

As you noticed, $P_n \to \frac{\pi}{2}$ since it's the inside of limit of the L.H.S of the Wallis Product Formula multiplied by $1$.  Since continuous maps preserve limits, this implies $\sqrt{2P_n} \to \sqrt{\pi} $ and note that 
$$\lim_{n \to \infty}\sqrt{2P_n} = \lim_{n \to \infty}\frac{\sqrt{2}(n!)^2 2^{2n}}{(2n)!\sqrt{2n+1 }} \\ = \lim_{n \to \infty} \frac{(n!)^22^{2n}}{(2n)!\sqrt{n+ \frac{1}{\sqrt{2}}}} \\ = \lim_{n \to \infty} \frac{(n!)^22^{2n}}{(2n)! \sqrt{n}}$$
Where the last equality was got by translation invariance.
