<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] I will show that
$2^{n(n+1)}
\gt (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}
$
for $n \ge 16$.
The right side is
$s(n)
=(n+1)^{n+1}\prod_{k=1}^{n} \left(\frac{n-k+1}{k} \right)^{n-k+1}
$.
$\begin{array}\\
s(n)
&=(n+1)^{n+1}\prod_{k=1}^{n} \left(\frac{n-k+1}{k} \right)^{n-k+1}\\
&=(n+1)^{n+1}\frac{\prod_{k=1}^{n}(n-k+1)^{n-k+1}}{\prod_{k=1}^{n} k^{n-k+1}}\\
&=(n+1)^{n+1}\frac{\prod_{k=1}^{n}k^k}{\prod_{k=1}^{n} k^{n-k+1}}\\
&=(n+1)^{n+1}\frac{\prod_{k=1}^{n}k^{2k}}{\prod_{k=1}^{n} k^{n+1}}\\
&=(n+1)^{n+1}\frac{\prod_{k=1}^{n}k^{2k}}{(n!)^{n+1}}\\
\end{array}
$
so, if 
$c=\frac12\ln(2\pi) 
\approx 0.92
$,
$\begin{array}\\
\ln(s(n))
&=(n+1)(\ln(n+1)-\ln(n!))+2\sum_{k=1}^{n}k\ln(k)\\
&\sim (n+1)(\ln(n+1)-\ln(n!))+2(\frac12  n^2\ln(n)-n^2/4+n\,\ln(n)/2)\\
&\sim (n+1)(\ln(n)+\frac1{n}-c-\frac12\ln(n)-n\ln(n)+n)+ n^2\ln(n)-n^2/2+n\,\ln(n)\\
&= \ln(n)(n+1-c-\frac12-n(n+1)+n^2+n)
+1+n(n+1)-n^2/2\\
&= \ln(n)(n+\frac12-c)
+(n^2+2n+2)/2\\
&= (n-.42)\ln(n)
+(n^2+2n+2)/2\\
\end{array}
$
If $t(n)
=2^{n(n+1)}
$,
$\ln(t(n))
=n(n+1)\ln(2)
$,
so we are done if
$(n-.42)\ln(n)+(n^2+2n+2)/2
< n(n+1)\ln(2)
$.
According to Wolfy,
this is true for
$n \ge 16$.
</s>