Prove by induction: for $n \ge 0$, $\frac{(2n)!}{n!2^n}$ is an integer Another prove by induction question:
for $n \ge 0$, $$\frac{(2n)!}{n!2^n}$$ is an integer
Base step: 
$$n = 0$$
$$\frac{(2 \times 0)!}{0! \times 2^0} = \frac{0!}{1 \times 1} = 1$$
Induction step: please help

Assume $\frac{(2n)!}{n!2^n}$ is an integer. $\frac{2(n+1)!}{(n+1)!2^{n+1}}=\frac{(2n+2)\cdot(2n+1)\cdot(2n)!}{(n+1)\cdot n!\cdot 2^n\cdot 2}=\frac{(2n)!}{n!2^n}\cdot \frac{(2n+2)(2n+1)}{2(n+1)}$. Notice that $\frac{(2n)!}{n!2^n}$ is an integer by the inductive assumption, and that $\frac{(2n+2)(2n+1)}{2(n+1)}=\frac{2(n+1)(2n+1)}{2(n+1)}=2n+1$.
