I'm having some trouble with this proof. Here's the question: Use mathematical induction and Euclid's Lemma to prove that for all positive integers s, if p and q

_{1},q

_{2},...,q

_{s}are prime numbers and p divides q

_{1}q

_{2}...q

_{s}, then p=q

_{i}for some i with 1 ≤ i ≤ s.

Here's what I know: Euclid's Lemma says that if p is a prime and p divides ab, then p divides a or p divides b. More generally, if a prime p divides a product a

_{1}a

_{2}...a

_{n}, then it must divide at least one of the factors a

_{i}. For the inductive step, I can assume p divides q

_{1}q

_{2}...q

_{s+1}and let a=q

_{1}q

_{2}...q

_{s}. Then, p divides aq

_{s+1}and either p divides a, p divides q

_{s+1}, or p=q

_{s+1}. I know that since q

_{i}is prime, it cannot divide q

_{i}unless p=q

_{i}for some 1 ≤ i ≤ s. I'm just not sure how to formulate the proof. Usually with Induction I can set some property P(n) and test it is true for some base like P(0) or P(1) for the base step. I'm unsure how to go about it here.