$$\begin{aligned} (a_k + 1)^k &= \left( a_k + \frac{1}{k-1} + \frac{1}{k-1} + \dots + \frac{1}{k-1} \right)^k \\ &\ge k^k \cdot a_k \cdot \frac{1}{(k-1)^{k-1}} \\ &= \frac{k^k}{(k-1)^{k-1}} \cdot a_k \end{aligned}$$
The inequality is strict unless $a_k = \frac{1}{k-1}$. Multiplying analogous inequalities for $k=2, 3, \dots, n$ yields
$$\begin{aligned} (a_2 + 1)^2 \cdot (a_3 + 1)^3 \cdots (a_n + n)^n &> \frac{2^2}{1^1} \cdot \frac{3^3}{2^2} \cdot \frac{4^4}{3^3} \cdots \frac{n^n}{(n-1)^{n-1}} \cdot a_2 \cdot a_3 \cdots a_n \\ &= n^n. \end{aligned}$$