prove/verify prime division $a_i$ positive integers for $1\le i\le n$
if $p$ prime and $p\mid a_1a_2\cdots a_n$ then $p\mid a_i$ for some $1\le i\le n$:
My thinking is to prove it by contraposition.  
$p$ does not divide $a_i$ for all $1\le i\le n$ implies $p$ does not divide $a_1a_2\cdots a_n$:
If $p$ does not divide $a_i$ for all $1\le i\le n$, $p$ being prime it cannot equal the product of any two or more therefore it cannot divide $a_1a_2\cdots a_n$. 
 A: The case $n=2$ is where all the work is; the rest is a trivial mathematical induction.  You need to prove that if $p$ is prime and $p\mid ab$ then $p\mid a$ or $p\mid b$.  That statement is called Euclid's lemma.  If you google that term, you should find something on it.
Notice that the assumption that $p$ is prime cannot be dropped: for example, $6\mid 4\cdot3$ but $6\nmid4$ and $6\nmid3$.
Look at this answer that I wrote almost three years ago.
It tells you that if $p\nmid a$ then $a$ has a multiplicative inverse mod $p$, i.e. a number $c$ such that $ac\equiv 1\bmod p$.  That means $ac=kp+1$ for some $k$.  It follows that $abc= kpb+b$.  But if $p\mid ab$ then $ab=\ell p$ for some $\ell$, so we have $\ell p c = kpb+b$.  This implies $b = \ell pc-kpb=p(\ell c - kb)$, so $p$ divides $b$.
A: $a_i$ positive integers for $1\le i\le n$
if $p$ prime and $p∣a_1a_2⋯a_n$ then $p∣a_i$ for some $1\le i\le n$:
$n=1$: (vacuously true)  $p$ prime and $p|a_1$ -> $p|a_1$
base case:
$n=2$: True by Euclid's Lemma.  $p|a_1a_2$ -> $p|a_1$ or $p|a_2$.
If $p|a_1$ finished.  If not , $p$ and $a_1$ relatively prime so exist $x,y$ so $px + a_1y = 1$.  Multiply by $a_2$ to get $pa_2x + a_1a_2y = a_2$.  If $p$ divides $pa_2$ and $a_1a_2$ then $p$ divides $a_2$.
$p(n) = p$ prime and $p∣a_1a_2⋯a_n\Rightarrow p∣a_i$ for some $1\le i\le n$
Assume true for n.
$p(n)$ true.
$p(n)\Rightarrow p(n+1)$:  $p|(a_1...a_n)a_{n+1}$:  If $p$ does not divide $a_1...a_n$, $p$ must divide $a_{n+1}$      
