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. 


Here is a proof by induction
that starts with my formula
for the right side.
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}
$
Therefore
$\begin{array}\\
\frac{s(n+1)}{s(n)}
&=\frac{(n+2)^{n+2}\frac{\prod_{k=1}^{n+1}k^{2k}}{((n+1)!)^{n+2}}}{(n+1)^{n+1}\frac{\prod_{k=1}^{n}k^{2k}}{(n!)^{n+1}}}\\
&=\frac{(n!)^{n+1}(n+2)^{n+2}(n+1)^{2(n+1)}}{((n+1)!)^{n+2}(n+1)^{n+1}}\\
&=\frac{(n!)^{n+1}(n+2)^{n+2}(n+1)^{n+1}}{((n+1)!)^{n+1}(n+1)!}\\
&=\frac{(n+2)^{n+2}(n+1)^{n+1}}{(n+1)^{n+1}(n+1)!}\\
&=\frac{(n+2)^{n+2}}{(n+1)!}\\
\end{array}
$
Since,
if 
$t(n)
=2^{n(n+1)}
$,
$\frac{t(n+1)}{t(n)}
=2^{(n+2)(n+3)-(n+1)(n+2)}
=2^{2(n+2)}
$,
if we can show that
$\frac{(n+2)^{n+2}}{(n+1)!}
< 2^{2(n+2)}
$,
we are done.
This is
$n+2
< 4((n+1)!)^{1/(n+2)}
$
or,
replacing $n+1$ by $n$,
$n+1
< 4(n!)^{1/(n+1)}
$.
Since
$n!
\gt (n/e)^n
$,
this is implied by
$\begin{array}\\
n+1
< 4(n/e)^{n/(n+1)}\\
or\\
(n+1)^{n+1}
< 4^{n+1}(n/e)^n\\
or\\
(n+1)e^n(1+1/n)^n
<4^{n+1}\\
or\\
(n+1)e^{n+1}
<4^{n+1}\\
or\\
n+1
< (4/e)^{n+1}\\
or,
\text{again replacing }n+1
\text{ by }n,\\
n 
< (4/e)^n\\
or\\
n^{1/n} 
< 4/e\\
\end{array}
$
Since
$n^{1/n}
\le e^{1/e}
< 1.45
$
and
$4/e
> 1.47
$,
this is true for all $n$.
Therefore
$\frac{s(n+1)}{s(n)}
< \frac{t(n+1)}{t(n)}
$
or
$s(n+1)
< t(n+1)\frac{s(n)}{t(n)}
$.
Therefore,
if
$s(n) < t(n)$,
$s(n+1) < t(n+1)$.
Since
$s(2)
=3^32^2\frac12
=54
$
and
$t(2)
=2^{6}
=64
\gt s(2)
$,
$t(n) > s(n)$
for
$n \ge 2$.
