$\gcd(a,b_1 \cdots b_k)=1$ if an only if $\gcd(a,b_i) = 1$ for $i = 1,\dots,k$ Suppose that $a,b_1,\dots,b_k$ are integers and I want to show that $\gcd(a,b_1 \cdots b_k)=1$ if and only if $\gcd(a,b_i) = 1$ for $i = 1,\dots,k$. In the direction of assuming $\gcd(a,b_1 \cdots b_k)=1$ to be true, would it be wrong to assume that as $a$ and $b_1 \cdots b_k$ have no common factors other than $1$ it is trivially true that for every $i$, $a$ and $b_i$ must not have any common factors other than $1$, otherwise $\gcd(a,b_1 \cdots b_k) > 1$? Similarly for the direction assuming $\gcd(a,b_i) = 1$ for $i = 1,\dots,k$ we have that for every $i$, $a$ and $b_i$ have no common factors other than $1$, so if we take any combination of multiplying the $b_i$ together, particularly that of all the $b_i$, $a$ and $\prod_i b_i$ must have no common factors other than $1$, otherwise $a$ and at least one of the $b_i$ must satisfy $\gcd(a,b_i) > 1$? 
Should I be proving both directions by contradiction? I am having a difficult time not seeing this as something that is just trivially true.
 A: Another way to show this is by Bezout's theorem: $\gcd(a,b)=1$ iff there are integers $x,y$ for which $ax+by=1.$ Now assuming $\gcd(a,b_1\cdots b_k)=1$ there are $x,y$ with 
$$ax+b_1\cdots b_k\cdot y=1.$$
By grouping this you can get integers showing $gcd(a,b_1)=1$, namely 
$$ax+b_1\cdot(b_2 \cdots b_k y)=1,$$
and similarly for $\gcd(a,b_r)=1$ for the larger $r$.
For the other direction, assume for each $r$ that $\gcd(a,b_r)=1$, so that there are integers $x_r,y_r$ for which
$$ax_r+b_ry_r=1.$$
Multiply all these together for $r=1,2,\cdots k$, and collect all products using the second summand only into $$b_1y_1b_2y_2\cdots b_ky_k=b_1\cdots b_k (y_1 \cdots y_k)$$
Thus the "Bezout coefficient" of the product of the b's is the product of the various y's. Now note that all the terms of the expansion other than those just used are divisible by $a$ so that, although the expression for it is cumbersome, there is an integer $X$ which, together with the above integer $Y=y_1 \cdots y_k$, give the desired equation $aX+(b_1 \cdots b_k)Y=1$, from which by Bezout we get $\gcd(a,b_1 \cdots b_k)=1.$
This doesn't use unique factorization, only the Bezout mechanics and algebra. I don't know if all algebraic systems satisfying the Bezout theorem have unique factorization.
A: HINT:
Consider $$F=A\cdot a+B\prod_{1\le i\le k}b_i$$ where $A,B$ are integers
If $(a,b_j)=d,$ clearly, $d$ divides $F$ for all  $A,B$
$\implies d$ will divide $(a,\prod_{1\le i\le k}b_i)$
A: I think the question is rather simple and I think your approach works. But maybe try to write things out more carefully. For instance, if $a, \prod\limits_{i=1}^n b_i$ have no common factors, can you formally show that this is true for each $(a, b_i)$? This is an easy statement, but it is good practice to use the definition of $x|y$ rather than to just conclude it is trivial.
I might point out that in some spaces, this isn't even true. For instance, in $\mathbb{Z} [\sqrt{-5}]$ (numbers of the form $a+b \sqrt{-5}$ for $a, b \in \mathbb{Z}$), one has $6=3 \cdot 2= (1+\sqrt{-5})(1- \sqrt{-5})$. In this case, the GCDs of $3, 1+ \sqrt{-5}$ and $3, 1- \sqrt{-5}$ are both 1 but the GCD of $3, 6$ is $3$.
