Problem proving inequality $\frac{{4}^{n}}{n+1}\le \frac{(2n)!}{(n!{)}^{2}}$

I skip the base case $n=0$ because it's obvious.

I know that this is very equivalent to: Prove by induction: $\frac{{2}^{2n}}{n+1}<\frac{(2n)!}{(n!{)}^{2}},n>1$

But I try to learn some tricks so maybe you can help me?

Assumption: $\frac{{4}^{n}}{n+1}\le \frac{(2n)!}{(n!{)}^{2}}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{4}^{n}\le \frac{(n+1)(2n)!}{(n!{)}^{2}}$ for some n.

Step: $n\to n+1$

${4}^{n+1}=\frac{(n+2)(2n+2)!}{((n+1)!{)}^{2}}$

I begin with manipulating the LHS:

${4}^{n+1}=4\cdot {4}^{n}\le ???$

$???\le \frac{(2n)!(2n+2)(2n+1)(n+1)}{(n!(n+1))}$

$=\frac{(n+1)(2n+2)!}{((n+1)!{)}^{2}}$

Note that I want to work my way from both sides to the mid where I get a very easy to check inequality by assumption.

Can you help me?

I skip the base case $n=0$ because it's obvious.

I know that this is very equivalent to: Prove by induction: $\frac{{2}^{2n}}{n+1}<\frac{(2n)!}{(n!{)}^{2}},n>1$

But I try to learn some tricks so maybe you can help me?

Assumption: $\frac{{4}^{n}}{n+1}\le \frac{(2n)!}{(n!{)}^{2}}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{4}^{n}\le \frac{(n+1)(2n)!}{(n!{)}^{2}}$ for some n.

Step: $n\to n+1$

${4}^{n+1}=\frac{(n+2)(2n+2)!}{((n+1)!{)}^{2}}$

I begin with manipulating the LHS:

${4}^{n+1}=4\cdot {4}^{n}\le ???$

$???\le \frac{(2n)!(2n+2)(2n+1)(n+1)}{(n!(n+1))}$

$=\frac{(n+1)(2n+2)!}{((n+1)!{)}^{2}}$

Note that I want to work my way from both sides to the mid where I get a very easy to check inequality by assumption.

Can you help me?