Show that for any integer $n\ge5$, the inequality ${2n \choose n}>3^n$ holds. I used induction, but got stuck at the last step. By cancelling common factors in the expanded $2n$ choose $n$, I got a simpler equation that didn't make sense.How do i prove this inequality?

The main step is to prove for $n+1$ then
$${2(n+1)\choose n+1}=\frac{(2n+2)!}{(n+1)!(n+1)!}=\frac{(2n+2)(2n+1)}{(n+1)^2}{2n\choose n}=\frac{2(2n+1)}{(n+1)}{2n\choose n}$$
Using the hypothesis
$${2(n+1)\choose n+1}>\frac{2(2n+1)}{(n+1)}\cdot 3^n$$
but
$$\frac{2(2n+1)}{n+1}=\frac{4n+2}{n+1}=\frac{3(n+1)+(n-1)}{n+1}=3+\frac{n-1}{n+1}>3$$
so,
$${2(n+1)\choose n+1}>\frac{2(2n+1)}{(n+1)}\cdot 3^n>3\cdot 3^n=3^{n+1}$$
