<s>[INST] Prove that $2^{n(n+1)}>(n+1)^{n+1}\left(\frac{n}{1}\right)^n\left(\frac{n-1}{2}\right)^{n-1}\cdots \left(\frac{2}{n-1}\right)^{2}\frac{1}{n}$ If $n$ be a positive integer $>1$, prove that
$$2^{n(n+1)}>(n+1)^{n+1}\left(\frac{n}{1}\right)^n\left(\frac{n-1}{2}\right)^{n-1}\left(\frac{n-2}{3}\right)^{n-2}\cdots \left(\frac{2}{n-1}\right)^{2}\frac{1}{n}$$
Please help me to prove the above. I have to use laws of inequality like AM-GM. But how to use it for this particular problem.
Edit:
Only use laws of inequality.

Edit 2
I want to solve this by using laws of inequality like weighted AM-GM. My attempt is the following 
Consider positive numbers $\left(\frac{n}{1}\right), \left(\frac{n-1}{2}\right), \left(\frac{n-2}{3}\right), \cdots \left(\frac{2}{n-1}\right), \frac{1}{n}$ with corresponding weights $n, n-1, n-2, \cdots 2,1$, respectively and applying weighted AM>GM, we get,
$$\frac{\left(\frac{n^2}{1}\right)+\left(\frac{(n-1)^2}{2}\right)+\left(\frac{(n-2)^3}{3}\right)+\cdots \left(\frac{2^2}{n-1}\right)\frac{1^1}{n}}{n+(n-1)+\cdots +2+1}>\left[\left(\frac{n}{1}\right)^n\left(\frac{n-1}{2}\right)^{n-1}\left(\frac{n-2}{3}\right)^{n-2}\cdots \left(\frac{2}{n-1}\right)^{2}\frac{1}{n}\right]^{\frac{n(n+1)}{2}}$$
I am unable to get the result because I am unable to get the sum $\left(\frac{n^2}{1}\right)+\left(\frac{(n-1)^2}{2}\right)+\left(\frac{(n-2)^3}{3}\right)+\cdots \left(\frac{2^2}{n-1}\right)\frac{1^1}{n}$
Please suggest me some possible approach. 

 [/INST] Notice that
$$
RHS = (n+1)^{n+1} \cdot 
\left(\frac{n}{1}\right) \cdot 
\left(\frac{n}{1}\cdot \frac{n-1}{2}\right) \cdots
\left(\frac{n}{1}\cdot \frac{n-1}{2} \cdots \frac{1}{n}\right) =
\prod_{k=1}^n \binom{n}{k} =
\prod_{k=0}^n \binom{n}{k}.
$$
Then, by AM-GM,
$$
RHS = (n+1)^{n+1} \cdot \prod_{k=0}^n \binom{n}{k} \le
(n+1)^{n+1} \cdot \left( \frac{\binom{n}{0}+\binom{n}{1}+\ldots+\binom{n}{n}}{n+1}\right)^{n+1} = LHS.
$$
Equality holds only if $\binom{n}0=\ldots=\binom{n}{m}$; i.e. for $n=1$.
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