If $a$ and $b$ are relatively prime then so are $a$ and $b^n$ for every positive integer $n$ I have been asked to prove the following via induction (as the textbook as suggested):

If $a$ and $b$ are relatively prime then so are $a$ and $b^n$ for every positive integer $n$

So, I did the following but I'm stuck at the induction step.
Base Case $n=1$
$(a,b^1) =(a,b)= 1$. This is true via the definition provided in the problem. 
Inductive Hypothesis
Suppose, this is true for all $n$. $$(a,b^n) = 1$$
Inductive Step for $n+1$
\begin{align}
&(a,b^{n+1}) = 1 \\
&(a,b^nb) = 1 \\
\end{align}
Now, I am stuck from here. Quite honestly, I think have done the entire induction wrong. what I would have done next is convert the $\gcd$ into the algebraic form of $ka + lb^nb = 1$ but I'm not too sure. 
Please provide only hints. 
Thanks a lot!
 A: Since you only want hints:
Try to prove the following identity:
$$\gcd(a,b\cdot c) = \gcd(a,b)\cdot\gcd(a,c)$$
This identity basically finishes your proof, given your induction hypothesis.
Also, regarding your induction hypothesis, you should change "true for all $n$" to "true for all $n\leq k$".  Otherwise, you are assuming what you want to prove.  After that little change, the inductive step is then to prove this is true for $n = k+1$.
A: Hope to provide a valuable new proof. The other one is very nice (+1) !
Inductive step: by hypothesis $(a, b^i)=1 $ for every $ i \leq k $ .
Suppose $(a, b^{k+1})> 1$. Let $ p $ be a prime such that $ p \mid a $ and $ p \mid b^{k+1} $. So $ p\mid b $ (definition of prime) so $ p \mid (a, b)=1 $. Absurd, so it must be $(a, b^n)=1 $ for every $ n $ natural. 
Another way (not inductive) is to write down the unique factorization of $ a $, $ b $ and $ b^n $ and trying to continue from here.
The first proof is a "very very light " induction, in other words induction is not so necessary here (in the sense we can prove the results easily without this technique). Anyway its better to know other approaches:) 
A: Inductive Step:
If $\gcd(a, b^n) = 1$ then $\forall \alpha, \beta\in \mathbb{I}: \alpha\beta  = a \rightarrow \left(\gcd(\alpha, b^n) = 1 \wedge \gcd(\beta, b^n) = 1\right)$.  Now let's try to attempt to prove that given the above and that $\gcd(a, b) = 1$, that $\forall \alpha, \beta \in \mathbb{I}: \alpha\beta = a \rightarrow \left(\gcd(\alpha, b^{n + 1}) = 1 \wedge \gcd(\beta, b^{n + 1}) = 1\right)$.
This seems like a candidate for proof by contradiction.  Let's assume the following: $\exists \alpha,\beta \in \mathbb{I}: \alpha\beta = a \rightarrow \left(\gcd(\alpha, b^{n + 1}) \neq 1 \vee \gcd(\beta, b^{n + 1}) \neq 1\right)$.  Right away this assumes we can find at least one value $\alpha \neq 1$ such that:
$$
a = \alpha\beta \wedge b^{n + 1} = \alpha\lambda b^n
$$
Where $\alpha\lambda = b$ and let's assume we have found such a value such that $\gcd(a, b^{n + 1}) = \alpha \neq 1$ (which would force that $\gcd(\beta, \lambda b^n) = 1$).  $\gcd(a, b^n) = 1$ is a given, therefore $\gcd(a, b^{n + 1}) = \gcd(\alpha\beta, \alpha\lambda) = \alpha$.  However, we assumed that $\gcd(\alpha\beta, \alpha\gamma) = \gcd(a, b) = 1$.  This contradicts the assumption that $\alpha \neq 1$, therefore if $\gcd(a, b^n) = 1$ and $\gcd(a, b) = 1$ then $\gcd(a, b^{n + 1}) = 1$.
