Prove that $a^n$ divides $b^n$ implies that $a$ divides $b$. 
For $n \ge 1,$ and positive integers $a,b,$ show the following:
$\rm (a)$ If $\gcd(a,b)=1,$ then $\gcd(a^n,b^n)=1.$
$\rm (b)$ The relation $a^n\,|\,b^n$ implies that $a\,|\,b.$

This is a problem from Elementary Number Theory by David M. Burton. I was able to solve part $\rm(a)$ by expanding $(ax+by)^{2n-1}$ using binomial theorem and factoring. To solve part $\rm(b)$, I used Bezout's theorem along with the following hint given in the book:

Put $d=\gcd(a,b)$ and write $a=rd,b=sd,$ where $\gcd(r,s) = 1.$ By part $\mathrm{(a)},\, \gcd(r^n,s^n)=1$. Show that $r=1$, whence $a=d.$ 

Before seeing the hint, I tried using mathematical induction as follows.

Let $S$ denote the set of all positive integers $n$ for which $a^n\,|\,b^n\implies a\,|\,b.$ We observe that when $n=1,$ $a\,|\,b\implies a\,|\,b.$ This means that $1\in S.$ 
Next, assume that $k\in S$ so that $a^k\,|\,b^k\implies a\,|\,b.$ Since $a^k\,|\,b^k,$ we can write $b^k$ as $$b^k=ma^k\tag1$$ for some integer $m$, and $b$ as $b=ra$ for some integer $r$. 
Multiplying $(1)$ by $b$ yields
  $$\begin{align*}
b^k\cdot b&=ma^k\cdot b,\\
\text{or,}\quad \quad b^{k+1}&=ma^k\cdot ra,\\
\text{or,}\quad\quad b^{k+1}&=mra^{k+1},\\
\text{or,}\quad\quad b^{k+1}&=ca^{k+a},\tag{where $c=mr$}\\
\text{i.e.,}\!\quad\quad a^{k+1}\,&|\,b^{k+1},
\end{align*}$$ which is precisely the case for $n=k+1,$ putting $k+1\in S$ whenever $k\in S$. According to the first principle of finite induction, $S$ must be the set of all positive integers.
$``\,\blacksquare"$

Is my proof correct?
 A: 
Next, assume that $k\in S$ so that $a^k\,|\,b^k\implies a\,|\,b.$ Since $a^k\,|\,b^k,$ we can write $b^k$ as $$b^k=ma^k\tag1$$ for some integer $m$, and $b$ as $b=ra$ for some integer $r$. 

There's a mistake here. What you know is that a certain implication (namely "$a^k\,|\,b^k\implies a\,|\,b$") is true, and that's all you know from the inductive hypothesis. That means that if you know $a^k \mid b^k$, then you know that $a \mid b$, but it doesn't tell you that the antecedent is true. So when you write "Since $a^k \mid b^k$," you're assuming something that you don't know to be true, and your proof goes wrong. 
This portion of the proof really should take the form

Next, suppose that $k\in S$ so that $a^k\,|\,b^k\implies a\,|\,b.$ We now need
  to show that $a^{k+1} \mid b^{k+1} \implies a \mid b$, i.e., that $k+1 \in S$. To show this, we assume that 
  $$
a^{k+1} \mid b^{k+1} \tag{*}
$$
  and will prove that $a \mid b$.

And then you have to write a proof, of course. Much of what you wrote is good stuff and can be recycled in some form. But that particular fragment --- "Since $a^k \mid b^k$" ---  was where your original proof went off the rails. 
A: Write the prime decompositions of $a$ and $b$ as $$a=\prod_{i=i}^tp_i^{\mu _i},b=\prod_{i=i}^tp_i^{\lambda _i},$$. where $\mu_i, \lambda_i \text { are non-negative integers for }1 \le i \le t.$ The condition $a^n | b^n$ is equivalent to $n \mu_i \le n\lambda_i,1 \le i \le t,$ which implies $ \mu_i \le \lambda_i,1 \le i \le t,$ so $a|b.$
