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. 

 A: 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 $n=1$.
A: 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
$n \ge 2$.
A: 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
$n \ge 16$.
A: Take $\log$ of both sides and verify that the inequality becomes equality for $n=1$. Take backward difference:
$$\nabla RHS=(n+1)\log (n+1)- n\log n+n\log n-\sum_{i=1}^n \log i\le 2n\log 2 =\nabla LHS \tag 1$$
holds for $n=1$. Take another backward difference $$\nabla^2 RHS=(n+1)\log(1+\frac 1 n)\le 2\log 2 =\nabla^2 LHS\tag 2$$
which holds for $n=1$ as well. Since $((x+1)\log(1+\frac 1 x))'=\log(1+\frac 1 x)-\frac 1 x<0$ on $x>0$, $(2)$ is a strict inequality for $n>1$ hence $(1)$ is a strict inequality for $n>1$ and hence the original (strict) inequality holds for $n>1$. Q.E.D.
ETA: The above also shows that $\frac {LHS}{RHS}\to 2\log 2\approx 1.386$ and possibly monotonically.
