Proof that ${{2n}\choose{n}} > 2^n$ and ${{2n + 1}\choose{n}} > 2^{n+1}$, with $n > 1$ i'm trying to proof these two terms. I started with an induction, but I got stuck...
Can anybody help?

Induction works pretty well, since:
$$\frac{\binom{2n+2}{n+1}}{\binom{2n}{n}}=\frac{4n+2}{n+1}=2\cdot\frac{2n+1}{n+1}\geq 3 $$
and:
$$\frac{\binom{2n+3}{n+1}}{\binom{2n+1}{n}}=2\cdot\frac{2n+3}{n+2}\geq 3$$
as soon as $n\geq 1$.
