How to prove that $\binom{n}{1}\binom{n}{2}^2\binom{n}{3}^3\cdots \binom{n}{n}^n \leq \left(\frac{2^n}{n+1}\right)^{\binom{n+1}{2}}$? 
How can we prove that $$\binom{n}{1}\binom{n}{2}^2\binom{n}{3}^3\cdots \binom{n}{n}^n \leq \left(\frac{2^n}{n+1}\right)^{\binom{n+1}{2}}$$

$\bf{My\; Try::}$ Using $\bf{A.M\geq G.M\;,}$ we get
$$\binom{n}{0}+\binom{n}{1}+\binom{n}{2} + \cdots+\binom{n}{n}\geq (n+1)\cdot \left[\binom{n}{0}\cdot \binom{n}{1}\cdot \binom{n}{2}  \cdots \binom{n}{n}\right]^{\frac{1}{n+1}}$$
So $$2^n\geq (n+1)\left[\binom{n}{1}\cdot \binom{n}{2}\cdots \binom{n}{n}\right]^{\frac{1}{n+1}}$$
How can I solve it after that? Help me.
Thanks.

Hint: You are close to the answer. Here is the right way: Apply the AM-GM:
$(a_1+2a_2+\cdots + na_n)^{\binom{n+1}{2}} \geq \binom{n+1}{2}^{\binom{n+1}{2}}a_1a_2^2\cdots a_n^n$, with $a_k = \binom{n}{k}$, and the left side is a popular sum that can be calculated by several methods one of which is using derivative of $(1+x)^n$
