Prove by induction: $2!\cdot 4!\cdot 6!\cdot\cdot\cdot (2n)!\ge ((n+1)!)^n$ Prove by induction: $2!\cdot 4!\cdot 6!\cdot\cdot\cdot (2n)!\ge ((n+1)!)^n$
For $n=1$ inequality holds.
$(*)$For $n=k$
$2!\cdot\cdot\cdot (2k)!\ge ((k+1)!)^k$
Multiplying LHS and RHS with $(2k+2)!$ gives
$$2!\cdot\cdot\cdot (2k)!(2k+2)!\ge ((k+1)!)^k(2k+2)!$$
Assume (by contradiction)$$2!\cdot\cdot\cdot (2k)!(2k+2)!< ((k+1)!)^k(2k+2)!$$
$$2!\cdot\cdot\cdot (2k)!(2k+2)!-((k+1)!)^k(2k+2)!<0$$
$$(2k+2)!(2!\cdot\cdot\cdot (2k)!-((k+1)!)^k)<0$$
$(2!\cdot\cdot\cdot (2k)!-((k+1)!)^k)\ge 0$ by $(*)$, thus inequality holds $\forall n\in\mathbb{N}$
Is this proof correct?

Define $S$ the wanted number. Then
$$
S=\prod_{i=1}^n (2i)!=\prod_{i=1}^n (2(n+1-i))!.
$$
Therefore, based on $a!b!\ge \max(a,b)!\ge \frac{a+b}{2}!$ whenever $a,b$ are integers with same parity, we conclude that
$$
S=\sqrt{\prod_{i=1}^n(2(n+1-i))!(2i)!}\ge \sqrt{\prod_{i=1}^n(n+1)!^2}=(n+1)!^n.
$$
