IMO, 2009

Assume on the contrary that $n$ divides $a_k(a_1 - 1)$. Then $n$ divides $a_i(a_{i+1} - 1)$ for $i \ge 1$ (where $a_{k+j} = a_j$); that is, $a_i \equiv a_i a_{i+1} \pmod{n}$ for all $i \ge 1$. It follows that
$$a_i \equiv a_i a_{i+1} \equiv a_i a_{i+1} a_{i+2} \equiv \dots \equiv a_i a_{i+1} \dots a_{i+k-1} \equiv a_1 a_2 \dots a_k \pmod{n}.$$
Therefore, we have $a_i \equiv a_j \pmod{n}$ for every pair of positive integers $i$ and $j$. On the other hand, $|a_i - a_j| < n$. We conclude that $a_i = a_j$, violating the given condition that $a_1, a_2, \dots, a_k$ are distinct.