Let $P(n)$ be the following implication: If $a, c, b_1, \ldots , b_n \in \Bbb Z$ Title says it all. Any help would be appreciated! 
Let $P(n)$ be the following implication: If $a, c, b_1, \ldots , b_n \in \Bbb Z$ such that gcd$(a, b_i) = 1$ for $i \in {1, . . . , n}$, and  $$a|c \prod_{i=1}^{n} b_{i},$$  then $a|c$. Prove by induction that $P(n)$ is true for all $n \in \Bbb N$.
 A: First lets prove that for $n = 1$, the statement is true. To do this let's walk through few lemmas.
 Before anything I expect you to know the following theorem
 if $(a,b) = \alpha$, then there exist two integers $x$ and $y$, such that $ax + by = \alpha$.
Lemma: $(a,b)=1$ and $(a,c)=1$ then $(a,bc)=1$
Proof: by the above assumed theorem we have $ax_1+by_1=1$ and $ax_2+by_2=1$. multiplying the two sides we can reach at a form like this $ax_3+bcy_3=1$.For more
Lemma: If $a|bc$ and $(a,b) = 1$ then $a|c$
Proof: $(a,b)=1$ then $ax+by=1$ 
since $a|bc$ means, we have $ak = bc$, then
$ayk = byc$, then
$ayk = (1-ax)c$, where $by = (1-ax)$ then
$ayk = c-axc$, finally
$a(yk+xc) = c$
Therefore $a|c$
Now we are ready to prove for the case $n=1$
$a|cb_1$ and $(a,b_1)=1$ then $a|c$ is true because of the above lemma
Now assume that the preposition is true for the case $n=k$.
Then for $n=k+1$
$(cb_1...b_k)b_{k+1} = c(b_1...b_kb_{k+1})$ 
and by the first lemma we have 
if $(a,b_1...b_k)=1$ and $(a,b_{k+1})=1$ implies $(a,b_1...b_kb_{k+1})=1$ 
Therefore $a|c \prod_{i=1}^{k+1} b_{i}$ implies
 $a|c \prod_{i=1}^{k} b_{i}$ because $(a,b_{k+1})=1$ and by the second lemma. Finally using our assumption we have $a|c \prod_{i=1}^{k} b_{i}$ implies $a|c$
