Assume $n>2$, $n\in\mathbb{Z}$, and $a_2,a_3,...a_n\in\mathbb{R}^+$ such that $a_2a_3\cdots a_n=1$. Prove: $$(1+a_2)^2(1+a_3)^3...(1+a_n)^n>n^n.$$
My attempt:
I have used AM-GM for inquality so the condition follows that
$$\frac{2^{n^2-2}}{a_2^{n-1}a_3^{n-2}...a_{n-1}^2a_n}>n^n,$$
but I do not know how to continue to a solution.
Please share your ideas in comments. Thanks!
For each $2 \leqslant k \leqslant n$,$$ 1 + a_k = \underbrace{\frac{1}{k - 1} + \cdots + \frac{1}{k - 1}}_{k - 1 \text{ many}} + a_k \geqslant k \left( \left( \frac{1}{k - 1} \right)^{k - 1} a_k \right)^{\frac{1}{k}}, $$ i.e.$$ (1 + a_k)^k \geqslant \frac{k^k}{(k - 1)^{k - 1}} a_k. $$ Thus$$ (1 + a_2)^2 \cdots (1 + a_n)^n \geqslant \frac{2^2}{1^1} × \cdots × \frac{n^n}{(n - 1)^{n - 1}} · a_2 \cdots a_n = n^n. $$
