Iranian Mathematical Olympiad

For the sake of contradiction, let $p_1, p_2, \dots, p_n$ be all the prime divisors of numbers in form of $a_i a_j a_k - 1$. Moreover, let $p$ be the smallest of these primes, and $M = \max(a_1, \dots, a_{n+1})$ (mind that $n$ is fixed.). Let $k$ be a positive integer number satisfying $p^k > M$. There are infinitely many $a_i$'s, therefore one satisfies the following properties:
1. $i > n + 2$
2. $a_i > (p_1 p_2 \dots p_n)^k + 1$
It's easy to see for every $j$, $1 \le l \le n$ exists such that $p_l^{k+1} \mid a_i a_j a_{n+2} - 1$.
Therefore, due to the pigeonhole principle, there exists a pair $1 \le m < t \le n+1$ such that for one $l$, $p_l^{k+1} \mid (a_i a_m a_{n+2} - 1, a_i a_k a_{n+2} - 1)$.
$$\implies p_l^{k+1} \mid a_i a_m a_{n+2} - 1 - (a_i a_k a_{n+2} - 1) = a_i a_{n+2} (a_m - a_k)$$
$$\implies p_l^{k+1} \mid (a_m - a_k)$$
$$\implies M > |a_m - a_k| \geq p_l^{k+1} > p^k > M$$
which is clearly a contradiction. Hence, the claim of the problem. ■