Let a and b be natural numbers. Use induction to prove that if gcd(a,b) = 1 then for all natural numbers n, gcd(a,b^n) = 1. Is my proof correct? Thank you!
Let $gcd(a,b)=1$. Suppose $gcd(a,b^n)=1$ for all $n\in\mathbb{N}$. This is proven by induction. 
Let $p(n): gcd(a,b^n)=1$. For the base case, we need to verify that $p(1)$ is true. If $n=1$ then $gcd(a,b^n)=gcd(a,b)=1$. Hence, $p(1)$ is true. 
For the inductive step, we must prove that $p(k)$ is true for some k implies $p(k+1)$ is true.
Suppose $p(k)$ is true for some k then $gcd(a,b^k)=1$. 
Now we want to show $gcd(a,b^{k+1})= gcd(a,b^kb)=1$.
By Bezout's identity, since there are non-zero integers $x,y,u,v$ so that ${ax+by=1}$ and ${au+cv=1}$, we have
$$
\begin{align}
bycv=(1-ax)(1-au)\\
=1-a(x+u-axu)\\
a(x+u-axu)+bcvy
=1.
\end{align}
$$
Therefore, $(a,bc)=1$ since it is a linear combination of $a$ and $bc$ that equals 1. If $a$ and $bc$ had a common factor, that common factor would divide any linear combination of $a$ and $bc$. 
Applying Bezout's Identity to our problem, we know since there are non-zero integers $x,y,u,v$ so that $ax+by=1$ and $au+b^kv=1$, we have $$byb^kv = (1-ax)(1-au)\\ =1-a(x+u-axu)\\$$
$$
a(x+u-axu)+b^kbvy = 1.$$ Thus, $(a,b^kb)=1$ since it is a linear combination of $a$ and $b^kb$ that equals 1. If $a$ and $b^kb$ had a common factor, that common factor would divide any linear combination of $a$ and $b^kb$. 
Therefore $gcd(a,b)=1, gcd(a,b^k)=1$. We know if $gcd(a,b)=1$ and $gcd(a,c)=1$ then $gcd(a,bc)=1$ by Bezout's identity. Therefore $p(k+1)$ is also true. Therefore, $p(n)$ is true for all natural numbers n by induction (i.e., $gcd(a,b^n)=1$ for all $n\in\mathbb{N}$). 

Your induction proof works. However, I think everything about $c$ is unnecessary. Just the part of your proof where you use $1 = ax + by$ and $1 = au + b^kv$ to get $1 = a(x+u-axu) + b^{k+1}(vy)$, and from that conclude that $\gcd(a, b^{k+1})$ is enough.
