<s>[INST] Prove $((n+1)!)^n < 2!\cdot4!\cdots(2n)!$ so I know I need to prove this via induction, but I am somewhat stuck. Here is what I have does so far.
Let $p(n) = (n+1)!^n \le 2!\cdot4!\cdot\ldots\cdot(2n)!$


*

*$p(2) = 3!^2\le 2!\cdot4!$

*Assume $p(n)$ is true.

*Prove $p(n+1)$

*$p(n+1) = (n+2)!^{n+1}\le 2!\cdot4!\cdot\ldots\cdot(2n+2)!$


$$\begin{align*}
&=(n+2)!^n\cdot(n+2)!\le\ldots\\
&=(n+1)!^n\cdot(n+2)^n\cdot(n+2)!\le\ldots
\end{align*}$$
Given that $(n+1)!^n\le 2!\cdot4!\cdot\ldots\cdot(2n)!$ (Using Inductive Hypothesis)
$$\begin{align*}
&=(n+2)^n\cdot(n+2)!\le(2n+2)!\\\\
&=(n+2)^n\le\frac{(2n+2)!}{(n+2)!}
\end{align*}$$
And I am stuck here. Any help anyone can give would be helpful.
 [/INST] Since you have reached $ (n+2)^n.(n+2)! < (2n+2)! $ , I will lead you from here.
From this we can observe that there are 'n' number of $(n+2)$ terms on the LHS.
we can easily conclude that:
$(n+2) < (n+3)$ ---> $1st  (n+2)$
$(n+2) < (n+4)$ ---> $2nd  (n+2)$
.......and so on till
$(n+2) < (n+n+2)$ ---> $nth  (n+2)$
So, we can clearly say that $(n+2)^n.(n+2)! < (n+3).(n+4)....(2n+2).(n+2)!$
=>$(n+2)^n.(n+2)! < (2n+2)!$
Hope the answer is clear !
</s>