For $n=2$, we have $a_1+a_2 = a_1 a_2$, which rewrites as $(a_1-1)(a_2-1) = 1$. We get $a_1 = 2, a_2 = 2$, so $(a_1^2 - 1)(a_2^2 - 1) = 9$.

$$|a_1 a_2 \dots a_n| = |a_1 + a_2 + \dots + a_n| \le |a_1| + |a_2| + \dots + |a_n|.$$
Dividing by $|a_1 a_2 \dots a_n| > 0$ we obtain
$$1 \le \frac{|a_1|}{|a_1 a_2 \cdots a_n|} + \frac{|a_2|}{|a_1 a_2 \cdots a_n|} + \dots + \frac{|a_n|}{|a_1 a_2 \cdots a_n|}.$$
If all the numbers have absolute value greater than or equal to $2$, then the right-hand side of the above inequality is less than or equal to $\frac{n}{2^{n-1}}$, i.e., $2^{n-1} \le n$.
For $n \ge 3$, we obtain (by induction or using Bernoulli's inequality) that $2^{n-1} > n$, which contradicts the previous inequality.
Therefore, at least one of the numbers has absolute value $1$, which implies $(a_1^2 - 1)(a_2^2 - 1)\dots(a_n^2 - 1) = 0$, a perfect square.

Remark. For every $n \ge 3$, there exist numbers satisfying the condition in the statement, for example
$$\underbrace{1+1+\dots+1}_{n-2 \text{ times}}+2+n = \underbrace{1 \cdot 1 \dots 1}_{n-2 \text{ times}} \cdot 2 \cdot n.$$