Assume the contrary. Then, the set of prime divisors of $\{a_n\}$ is finite (and of course $\{a_n\}$ is an infinite set since $a_{n+1} > a_n$ for $n\geq 2$). But then, by Kobayashi, the set of prime divisors of $\{a_n+1\}$ is infinite. But $a_n +1 = a_1a_2\cdots a_{n-1}$, whose prime divisors must belong to the original set, a contradiction.

Well, of course Kobayashi is a complete overkill here: Once a prime number divides an element of the sequence, it will never divide anything in the sequence again, so we must even have a new prime factor in each step.

Observe that for prime $p$, if $p | a_n$ then $p \nmid a_k \forall k > n$. Therefore, we must get a new prime divisor for each $a_k$, proving the result. $\square$ (Remark: One also has to show that $a_i > 1 \forall i$, but that's easy.)