Let $p$ be a prime number and $a_1, a_2, \ldots, a_n$ be integers. Prove that if $p \mid a_1a_2 \ldots a_n$, then $p \mid a_j$ for some $j$ with $1 \leq j \leq n$.
The hint was to use induction.
So for the base case ($n = 1$):
$p \mid a_1$ would imply $p \mid a_j$ for some $j$ with $1 \leq j \leq 1$, so $j$ would have to equal 1, which would say that $p \mid a_1$ which was assumed so the base case holds.
For the induction step:
I would assume that if $p \mid a_1a_2 \ldots a_k$ then it is true that $p \mid a_j$ for some $j$ with $1 \leq j \leq k$, and would have to prove that $p \mid a_1a_2 \ldots a_k \cdot a_{k + 1}$ implies that $p \mid a_j$ for some $j$ with $1 \leq j \leq k + 1$.
This is where I get stuck, I think you only have to check the case where $j = k + 1$ because if $j$ was $a_1$ through $a_k$ the initial condition would hold? But I'm not sure.