Show that that $\lim_{n\to\infty}\sqrt[n]{\binom{2n}{n}} = 4$

I know that

$$ \lim_{n\to\infty}{{2n}\choose{n}}^\frac{1}{n} = 4 $$ but I have no Idea how to show that; I think it has something to do with reducing ${n}!$ to $n^n$ in the limit, but don't know how to get there. How might I prove that the limit is four?

Hint: By induction, show that for $n\geq 2$ $$\frac{4^n}{n+1} < \binom{2n}{n} < 4^n.$$

The term ${2n\choose n}$ occurs as biggest of $2n+1$ positive summands in expanding $(1+1)^{2n}$, which directly shows $\frac{4^n}{2n+1}\le {2n\choose n}\le 4^n$. From this the claim follows by using $\sqrt[n]n\to 1$.

Use Stirling formula $$n! \sim \sqrt{2 \pi n} \left(\dfrac{n}e \right)^n$$ This gives us that $$\dbinom{2n}n \sim \dfrac{4^n}{\sqrt{\pi n}}$$ Now conclude that $$\lim_{n \to \infty} \dbinom{2n}n^{1/n} = 4$$