For all n > 0 prove that ${2n}\choose{n}$ < $4^n$
It seems like a simple proof but I'm not sure how to continue it.
What I have so far:
Base Case:
Let n=1 ${2}\choose{1}$ = 2
$4^1$ = 4
2 < 4
Inductive Step: Assume ${2(x+1)}\choose{x+1}$ < $4^{x+1}$
Left side: $\frac{(2x+2)!}{(x+1)1(x+1)!}$
For the inductive step:
$$ \begin{align*} \binom{2n+2}{n+1} &= \frac{(2n+2)!}{(n+1)!\,(n+1)!} \\ &= \frac{(2n+2)(2n+1)}{(n+1)^2} \times \frac{(2n)!}{n!\,n!} \\ &= \frac{(2n+2)(2n+1)}{(n+1)^2} \binom{2n}{n} \\ &= \frac{2(2n+1)}{n+1}\binom{2n}{n} \\ &= \frac{4n + 4 - 2}{n+1}\binom{2n}{n} \\ &= \left(4 - \frac{2}{n+1}\right)\binom{2n}{n} \\ &< 4\binom{2n}{n} \\ &< 4^{n+1}. \end{align*} $$