TITLE: Product of pairwise coprime integers divides $b$ if each integer divides $b$
QUESTION [8 upvotes]: Let $a_1....a_n$ be pairwise coprime.  That is $gcd(a_i, a_k) = 1$ for distinct $i,k$, I would like to show that if each $a_i$ divides $b$ then so does the product.
I can understand intuitively why it's true - just not sure how to formulate the proof exactly.
I want to say if we consider the prime factorizations of each $a_i$, then no two prime factorizations share any prime numbers.   So the product of $a_1...a_n$ must appear in the prime factorization of $b$.  Is this correct?  Or at least if if the idea is correct, any way to formulate it more clearly?

REPLY [12 votes]: One way of arguing that avoids looking at prime factorizations is to note that $gcd(a,b)=1$ iff $a\alpha+b\beta=1$ for some integers $\alpha,\beta$.
But then, if $a,b$ both divide $c$, say $at=c$ and $br=c$ where $t,r$ are integers, we have
 $$ ab(\alpha r+\beta t)=br(\alpha a)+at(\beta b)= c(\alpha a+\beta b)=c. $$
This explicitly shows that $ab$ also divides $c$. Now use induction.