Knowing that $\prod_{i = 1}^na_i = 1$, prove that $\prod_{i = 1}^n(a_i + 1)^{i + 1} > (n + 1)^{n + 1}$. 
Given natural $n$ $(n \ge 3)$ and positives $a_1, a_2, \cdots, a_{n - 1}, a_n$ such that $\displaystyle \prod_{i = 1}^na_i = 1$, prove that $$\large \prod_{i = 1}^n(a_i + 1)^{i + 1} > (n + 1)^{n + 1}$$

We have that $$\prod_{i = 1}^n(a_i + 1)^{i + 1} \ge \prod_{i = 1}^n(2\sqrt{a_i}) \cdot \left(\sqrt[m]{\prod_{i = 1}^na_i^i} + 1\right)^m$$
where $\displaystyle p = \sum_{i = 1}^ni = \dfrac{n(n + 1)}{2}$, then I don't know what to do next.
I suspect that $\displaystyle \min\left(\prod_{i = 1}^n(a_i + 1)^{i + 1}\right) = 2^q$, occuring when $a_1 = a_2 = \cdots = a_{n - 1} = a_n = 1$, where $q = \dfrac{(n + 3)n}{2}$, although I'm not sure that $2^q > (n + 1)^{n + 1}, \forall n \in \mathbb Z^+, n \ge 2$.
(I've just realised this is just a redraft of problem 2, IMO 2012.)

There were a few details in Le Thanh Dat's solution that were not clear to me, so I offer these extra details.
Rewrite each bracket 
\begin{eqnarray*}
a_i+1 = \frac{1}{i+1} \left( (i+1)a_i + \underbrace{ \frac{i+1}{i} +  \cdots + \frac{i+1}{i}}_{ \text{$i$ terms}} \right).
\end{eqnarray*}
We can now apply AM-GM 
\begin{eqnarray*}
\prod_{i=1}^{n} (a_i+1)^{i+1} \geq \prod_{i=1}^{n} a_i \frac{(i+1)^{i+1}}{i^i} = (n+1)^{n+1}.
\end{eqnarray*}
The last equality follows using $\prod a_i = 1$ and telescoping.
