I came up with this proof for the inequality. If the proof is correct, is it possible to come up with a shorter one? I feel like my proof is very inelegant and there is a short trick that can be applied.
Prove the following by induction (B. Demidovich, Problems in Mathematical Analysis, §1, ex. 9):
$$ \prod_{k=1}^{n} {(2k)! \gt ((n+1)!)^n, \ \ n\gt 1} \ \ \ (a)$$
Base case for n = 2 is trivial and boils down to: $$ (2!\cdot4!) = 48 \gt 36 = (3!)^2$$
Assuming (1) is valid for n, we need to prove:
$$\prod_{k=1}^{n+1} {(2k)! \gt ((n+2)!)^{n+1}, \ \ n\gt 1} \ \ \ (b) $$
I will rewrite the statement as:
$$(2n+2)!\cdot\prod_{k=1}^{n} {(2k)! \gt ((n+2)!)^{n+1}, \ \ n\gt 1}$$
Using hypothesis (a) we see that it is enough to prove:
$$(2n+2)!\cdot((n+1)!)^n\gt ((n+2)!)^{n+1}$$
Let's perform equivalent transformaiton on the second term of the left side:
$$ (2n+2)!\cdot(\frac{(n+2)!}{n+2})^n\gt ((n+2)!)^{n+1} $$
Taking out $(n+2)$ from outside of brackets:
$$ \frac{(2n+2)!}{{(n+2)^n}}\cdot({(n+2)!})^n\gt ((n+2)!)^{n+1} $$
It is left now to prove that:
$$ \frac{(2n + 2)!}{{(n+2)^n}}\gt (n+2)! \ \ \ $$
Rewritten like this:
$$\frac{(n+2)!\cdot\prod_{k=1}^{n} {((n+2)+k)}}{\prod_{k=1}^{n} {(n+2)}} \gt (n+2)! $$
Eliminating $(n+2)!$ on both sides leaves us with:
$$\prod_{k=1}^{n} {((n+2)+k)} \gt \prod_{k=1}^{n} {(n+2)} $$
Which clearly holds due to $1 \le k \le n $. Therefore (b) holds. ∎
Write like this
$$(2n+2)!\cdot\prod_{k=1}^{n} {(2k)! \gt ((n+2)!)^{n}(n+2)!}=((n+1)!)^{n}(n+2)^{n}(n+2)!$$
Since by hypothesis
$$\prod_{k=1}^{n} {(2k)! \gt ((n+1)!)^{n}}$$
You eliminate(see edit) those and check
$$\frac{(2n+2)!}{(n+2)!}\gt (n+2)^{n}\Longrightarrow (n+3)\cdots(2n+2) \gt (n+2)^{n}$$
What is clearly true since each term at the LHS is bigger than $(n+2)$.
EDIT: just to be clear that here "eliminates" doesn't mean to cancel out factors, it means to assume a certain property so that we can check the remaining inequality. If $a>d$ and $b>c$ and all of them are positive$^1$, we can conclude $ab>cd$ by transitivity:
$$ a>d\Longrightarrow ab>bd $$ $$ b>c\Longrightarrow bd>cd $$ $$ ab>bd>cd\quad \blacksquare $$
But from $ab>cd$ and $b>c$, we cannot conclude $a>d$. Just put $a=d$ in the first inequality.
$^1$ We also cannot use the transitivity above when factors' multiplication might change signal. Put $1>-1$ and $-2>-3$, clearly $-2>3$ is false.