For $n>1$, estimate the product $(n+1)^{n+1}(n/1)^n\dots (1/n)$ from above If $ n $ be a positive integer $>1$, prove that
$$2^{n(n+1)}\gt(n+1)^{n+1}\biggl(\frac{n}{1}\biggr)^{n}\biggl(\frac{n-1}{2}\biggr)^{n-1}...\biggl(\frac{2}{n-1}\biggr)^{2}\biggl(\frac{1}{n}\biggr)$$
 A: Consider $(n+1)$ positive numbers $\binom{n}{0}, \binom{n}{1}, ...\binom{n}{n}$. $$ A.M.= \frac{\binom{n}{0}+ \binom{n}{1}+ ...+\binom{n}{n}}{n+1}=\frac{(1+1)^n}{n+1}=\frac{2^n}{n+1}$$
$$G.M.=\sqrt[(n+1)]{\binom{n}{0} \binom{n}{1}. ...\binom{n}{n}}=\sqrt[(n+1)]{\biggl(\frac{n}{1}\biggr)^n \biggl(\frac{n-1}{2}\biggr)^{n-1} \biggl(\frac{n-2}{3}\biggr)^{n-2}...\biggl(\frac{2}{n-1}\biggr)^{2} \biggl(\frac{1}{n}\biggr)^{1}} $$  Applying   A.M.> G.M., we get $$ \biggl(\frac{2^n}{n+1}\biggr)^{n+1}> \biggl(\frac{n}{1}\biggr)^n \biggl(\frac{n-1}{2}\biggr)^{n-1} \biggl(\frac{n-2}{3}\biggr)^{n-2}...\biggl(\frac{2}{n-1}\biggr)^{2} \biggl(\frac{1}{n}\biggr)^{1}$$
A: This is too long to be a comment:You can simplify the RHS to be one of the following based on the parity of $n$. $$(n+1)^{(n+1)}\prod_{i=1}^{\frac{n-1}{2}}\left(\frac{n-i}{i}\right)^{2i+1}\quad\text{or}\quad
(n+1)^{(n+1)}\left(\frac{n}{2}\right)^\frac{n}{2}\prod_{i=1}^{\frac{n-2}{2}}\left(\frac{n-i}{i}\right)^{2i+1}$$
clearly the one on the right is much bigger, and so you should work from there, but it's not at all clear to me that that one actually satisfies this inequality.
