Prove that if $p$ is prime and $p\mid a_1 a_2\cdots a_n$ then $ p| a_i$ for some $ i = 1, 2, ...., n$. Question:
Prove that if $p$ is prime and $p\mid a_{1} a_{2}\cdots a_{n}$ then $ p\mid  a_{i}$ for some $ i = 1, 2, ...., n$.
My attempt:
Proof by Induction:
Base Case: $n=1$, we have $ p\mid a_{1} \implies p\mid a_{1} $. So Base case holds.
I.H: Assume $ \ p\mid a_{1} a_{2}\cdots a_{k} \implies p\mid a_{i}\ $ for some  $\ i = 1, 2, ... k$ 
Consider $ n = k+1$
$ p\mid a_{1} a_{2}\cdots a_{k+1} \implies p\mid (a_{1} a_{2} \cdots a_{k})\, a_{k+1} \implies p\mid a_{1} a_{2}\cdots a_{k}\, $ or $ \ p\mid a_{k+1}$
If $\  p\mid a_{k+1}$ then we are done. Else by I.H $ \ p\mid a_{1} a_{2} \cdots  a_{k} \implies p\mid a_{i}\ $ for some $\ i = 1, 2, ... k$ 
 A: Your proof is fine. In fact we can similarly inductively extend any property $P$ that satisfies $\ P(ab) = P(a)\vee P(b)\,$ to products of any length, where $x \vee y := x\,$ or $\,y.\,$ Namely 
$$\begin{align} P((a_1\cdots a_n) a_{n+1})\, &= \qquad\ \ \, P(a_1\cdots a_n)\vee P(a_{n+1})\\[.3em]
&=\, P(a_1)\vee \cdots P(a_n)\vee P(a_{n+1})\ \ {\rm by\ \ induction}
\end{align}$$
In your case $\,P(a) := p\mid a.\,$ Note that the only property used of multiplication and $\vee$ is associativity, so the proof is really about $n$-ary extension of monoid homomorphisms.
A: Now let me do it, considering you meant product. I will use $P$ to refer from here on to $a_1 a_2 ... a_n$. Note that $p|P$, so $P$ contains at least 1 factor of $p$. When you now partition $P$ as a product of some integers, $p$ must obviously show up somewhere in that product. Either one of the $a_i$ is therefore $p$, or one of the $a_i$ contains $p$ in its prime decomposition. If one $a_i$, say $i=x$ equals $p$, then clearly $a_x/p=p/p=1 \in \mathbb{Z} \implies p|a_x$, which satisfies what you originally wanted to prove (Q.E.D). Or you could have for a certain $a_i$, once again, say $i=x$, that contains a factor of $p$ in its prime decomposition. Therefore, $a_x=p b$, for some integer $b$. This means that $a_x/p=b \in \mathbb{Z} \implies p|a_x$ [Q.E.D.].
A: We divide this into two main cases, either all $a_i$ are relatively prime, or they are non-relatively prime. First, for the relatively prime case, clearly $p=1$, and therefore $p$ must evenly divide all $a_i$. Now consider the second case. Let the GCD be $g$, where $g$ is not $1$. Now, for all $a_i$ let $a_i=gb_i$, where all $b_i$ are integral. We know $p$ divides $g$, so $p$ is a prime factor of $g$. Therefore, $g=\alpha p^\lambda$, for integers $\alpha,\lambda$, so for all $a_i$, $a_i=\alpha p^\lambda b_i$, and $a_i/p=\alpha p^{\lambda -1} b_i \in \mathbb{Z} \implies p|a_i$, for all $i$ [Q.E.D]
EDIT: This proof was done assuming you meant GCD. I will keep it on here for educational purposes anyway, though. I redid the proof using the fact that you meant product instead, so refer to that.
A: This step 
$p\mid a_{1} a_{2}\cdots a_{k+1} \implies p\mid (a_{1} a_{2} \cdots a_{k})\, a_{k+1} \implies p\mid a_{1} a_{2}\cdots a_{k}\,$ or $\ p\mid a_{k+1}$.
assumes precisely what you want to prove:  That if $p\mid Ma$ then $p\mid M$ or $p \mid a$.
If you are going to assume this you must make your base case $k=2$ and not $k = 1$ else you will have an all horses are the same color error.
Case $k=2$ is precisely Euclid's Lemma
If you know Bezout's Lemma we know that if $p|a_1a_2$ and $p\not \mid a_1$ then we know there are integers $m,n$ so that $pm + na_1 = 1$  Therefore $pma_2 + na_1a_2 = a_2$.  But $p|a_1a_2$ and $p|pma_2$ so $p|pma_2 + na_1a_2 = a_2$.  So either $p|a_1$ or if not, $p|a_2$.
NOW your induction step works just fine.
