Ok this is a little confusing, so suppose for all $n \geq 2$, $a_1 a_2a_3...a_n \mid k$, then how do I show $a_1 \mid k \land a_2 \mid k \land ... a_m \mid k$, and the reverse, if $\gcd(a_1,a_2,a_3...,a_n) = 1$? That is their only common divisor is 1, and k is any integer.
This certainly looks like a proof by induction. I should need to prove the induction hypothesis instead, that is $P_m => P_{m+1}$ for some m. But I'm stuck at the following steps:
I let $k = b\cdot a_1 ...a_{m+1}$ at the start, and extract $b\cdot a_1 ...a_{m}$ out of the expression and so show that $a_{m+1} | k$
Then since $a_1 \mid k \land a_2 \mid k \land ... a_m \mid k$ is true, adding one more term which is true (i.e. $a_{m+1} \mid k)$ will give true also.
But this doesn't really use the information that the variables have no common divisors. Suppose $a_1$ to $a_n$ do not have any common divisors, does this reliably mean that all of them are primes?
There are two things to prove here.
That $a_1a_2\cdots a_n$ divides $k$ implies that each $a_i$ divides $k$. Your proof by inductions works fine for this (but I think there's a shorter proof.) Note that here you don't have the hypothesis that the gcd is 1, which is why you didn't use it.
That, with the additional hypothesis that the gcd is 1, then the product divides $k$. As was pointed out, this is not a sufficient hypothesis. There is a difference between $\gcd(a_1, a_2,\ldots,a_n) = 1$ and the phrase "the $a_i$ are pair-wise relatively prime." You can see this in Lord Shark's example. To prove the "reverse" (most people would call it "converse") you need "pair-wise relatively prime" which means that $gcd(a_i, a_j) = 1$ for every choice of $i$ and $j$. Again, induction will work here just fine. Bezout's lemma is needed at the key step.