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.
For 4, $p(n+1)$ is a statement, not a term, so should not be set equal to something. The $=$ should be logical equivalence. You are trying to prove $p(n) \implies p(n+1)$ This is a nit with your presentation, but is an important concept.
Under 4, you have $(n+1)^n \le \prod_{i=1}^{n}(2i)!$ and want to prove $(n+2)^{n+1} \le \prod_{i=1}^{n+1}(2i)!$. Dividing them we want to prove $\left(\frac {n+2}{n+1} \right)^n(n+2)\le(2n+2)!$ The first paren on the left is less than $e$ so if you have $n$ large enough you are there.