Part 1: By Cauchy-Schwarz (or the QM-AM inequality), $\frac{\binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2 + \cdots + \binom{n}{n}^2}{n} > \frac{\left(\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n}\right)^2}{n}$
so $\binom{2n}{n} > \frac{(2^n)^2}{n} > 2^n$.
Part 2: $\binom{2n}{n} = \frac{2^n(1\cdot 3\cdot 5\cdots(2n-1))}{n!} < \frac{2^n \cdot n^n}{n!} = \frac{2^n}{n!/n^n} = \frac{2^n}{\prod_{j=0}^{n-1}\left(1-\frac{j}{n}\right)}$.