Have fun with these expressions!

Fix an integer $k$ such that $2 \leq k \leq n$ . We start by applying weighted AM-GM inequality to the positive numbers $\frac{1}{k-1}$ and $a_k$ with weights $k-1$ and $1$ respectively:

$\frac{1+a_k}{k} = \frac{(k-1)\frac{1}{k-1}+a_k}{k} \geq ((\frac{1}{k-1})^{k-1} \cdot {a_k}^1)^{\frac{1}{k}} = (\frac{a_k}{(k-1)^{k-1}})^{\frac{1}{k}}$

Therefore it follows that $(1+a_k)^k \geq \frac{k^k}{(k-1)^{k-1}}a_k$ for every $2 \leq k \leq n$ . Equality holds when $a_k=\frac{1}{k-1}$ .

Multiplying together these inequalities we obtain:

$a_1(1+a_2)^2(1+a_3)^3 \ldots (1+a_n)^n \geq a_1a_2a_3 \ldots a_n \cdot \frac{2^2}{1^1} \cdot \frac{3^3}{2^2} \cdot \ldots \cdot \frac{n^n}{(n-1)^{n-1}} = a_1a_2a_3 \ldots a_n \cdot n^n = n^n$

Equality occurs only when $a_k=\frac{1}{k-1}$ for every $2 \leq k \leq n$ .

Since we did not use the value of $a_1$ in this reasoning, we may assume that $a_1=(n-1)!$ so that the condition $a_1a_2 \ldots a_n=1$ is met. Thus we have shown that the conditions are fulfilled only by these values of $a_1, \ldots, a_n$ , hence the value of our desired expression is easy to determine now:

$\frac{a_1}{n-1}(\frac{1}{a_2} + \frac{1}{a_3} + \ldots + \frac{1}{a_n}) = \frac{(n-1)!}{n-1}(1 + 2 + \ldots + n-1) = (n-2)! \cdot \frac{(n-1)n}{2} = \frac{n!}{2}$ .