TITLE: From $n=2$ to $n$-ary Euclid lemma $\ p\mid a_1\cdots a_n\Rightarrow\ p\mid a_1\,$ or $\,\cdots\, p\mid a_n$
QUESTION [5 upvotes]: How to prove a generalized Euclid lemma par induction after proving Euclid lemma?
I want to prove the generalized lemma, to prove by rearranging the product of number and use Euclid lemma as a model. A proof will be nicer if it can use induction principle.

REPLY [4 votes]: The induction almost writes itself: it is just a matter of writing down your informal reasoning in the language of induction. 
We want to prove that for every integer $n\ge 2$, if the prime $p$ divides the product $a_1a_2\cdots a_n$, then $p$ divides at least one of the $a_i$.  The result has been already proved for $n=2$. It remains to do the induction step.
Suppose the result is true when $n=k$. We show the result is true when $n=k+1$.
So we are told that $p$ divides $a_1a_2\cdots a_ka_{k+1}$. Thus $p$ divides $(a_1a_2\cdots a_k)a_{k+1}$. This is a product of two numbers. Thus $p$ divides $a_1a_2\cdots a_k$ or $p$ divides $a_{k+1}$. 
If $p$ divides $a_{k+1}$ we are finished. If $p$ does not divide $a_{k+1}$, then $p$ divides $a_1a_2\cdots a_k$. Then by the induction hypothesis $p$ divides one of the $a_i$ ($1\le i\le k$), and we are finished.