Another approach that I thought of :
Set $a_2 =\frac{x_2}{x_3}, a_3=\frac{x_3}{x_4},\ldots, a_n=\frac{x_n}{x_2}$. This is a very useful substitution that we use in cases when we have a product equal to one like in this one $ a_2 a_3 \cdots a_n=1 $.
Now we need to prove that $ (x_2+x_3)^2 (x_3+x_4)^3 \cdots (x_n+x_2)^n > n^n x_3^2 x_4^3 \cdots x_{n}^{n-1}x_2^n$
which become obvious since for each $k$ by applying the Arithmetic-Geometric Mean we have that: $ (x_k+x_{k+1})^{k}=\left(x_k+(k-1)\frac{x_{k+1}}{k-1}\right)^k\geqslant k^k x_k\frac{x_{k+1}^{k-1}}{(k-1)^{k-1}} $
Just multiply for $k$ from $2$ to $n$.