The relation $a^n|b^n$ implies that $a|b$, where $a$ and $b$ are positive integers and $n\ge1$. Let $gcd(a, b) = d$
Let $a = rd$ and $b = sd$, where $gcd(r, s) = 1$.
$\therefore gcd(r^n, s^n) = 1$
Now, how do I show that $r = 1$?
 A: You're on the right track:
$b^n = a^n t \implies s^n = r^n t \implies r \mid s^n \implies r = 1$ because $\gcd(r,s)=1$.
Another way
$a^n\mid b^n$ iff $v_p(a^n) \le v_p(b^n)$ for all primes $p$. Here, $v_p(x)$ is the exponent of $p$ in the prime factorization of $x$.
Now, $v_p(a^n) \le v_p(b^n)$ iff $n v_p(a) \le n v_p(b)$ iff $v_p(a) \le  v_p(b)$ iff $a \mid b$.
A: Okay.  It shouldn't be hard to prove $\gcd(m^k, n^k) = [\gcd(m,n)]^k$.
You can do it by considering prime factors, or by considering $$\gcd(m^k,n^k)=\\\gcd(m^{k-1}\frac m{\gcd(m,n)}\gcd(m,n), n^{k-1}\frac n{\gcd(m,n)}\gcd(m,n))=\\\gcd(m,n)\gcd(\frac m{\gcd(m,n)}m^{k-1},\frac m{\gcd(m,n)}n^{k-1})=\\\gcd(m,n)\gcd(m^{k-1},n^{k-1})$$ and using induction.
Or we could do it your way.
If $\gcd(m,n) = d$ and $m=rd;n=sd$ with $\gcd(r,s) =1$. then $\gcd(m^k,n^k)=\gcd(d^kr^k,d^ks^k)=d^k\gcd(r^k,s^k) = d^k*1 = d^k$.
.......
So now bear in mind $g|h \iff \gcd(g,h) = g$.  (That's too obvious to require any discussion isn't it?  If $g$ is a common divisor $g$ and $g$ then $g|h$.  ANd if $g|h$ than $g$ is a common divisor and there can't be any greater divisor of $g$ than $g$ itself... right?)
So we are done.
......
$a^n|b^n\implies a^n = \gcd(a^n,b^n) = d^n\gcd(r^n,s^n) =d^n*1 =d^n\implies a=d=\gcd(a,b) \implies a|b$.
That's it.
