Proof for Perfect $n^{th}$ power.

Question

A positive integer, $k$, is a perfect nth power if, there exist a positive integer a such that $k=a^n$, where $n$ is also a positive integer.

Question is:
How can I prove, for all positive integers $k, m, n$ (where $m$ and $n$ are coprime), if $k$ is a perfect $m^{th}$ power and a perfect $n^{th}$ power, then it is a perfect $mn^{th}$ power. This looks a lot like perfect squares, but it is more than perfect squares. So for example $16 = 2^4 = 4^2$

I'm not sure how to do this. I tried some patterns, is it possible to use gcd? I know that if $gcd(a, b)=1$, then $gcd(a^n, b^n)=1$. But I don't know how this can help either.

Answer 1

Hint: $(m,n)=1 \Rightarrow \exists x, y \in \mathbb{N}, mx-ny=1.$

Can you start from here?

If $k=a^m=b^n$, then $a=a^{mx-ny}=\left(\frac{b^x}{a^y}\right)^n.$

Answer 2

If you know the fundamental theorem of arithmetic, then you should be able to show the following lemma:

A positive integer $k$ is a perfect $n$-th power if and only if for every prime number $p$, the exponent of $p$ appearing in the factorization of $k$ is a multiple of $n$.

Now if $k$ is a perfect $m$-th power and a perfect $n$-th power, then by the lemma, for every prime $p$, the exponent of $p$ appearing in the factorization of $k$ is simultanously a multiple of $m$ and a multiple of $n$.

Since $m$ and $n$ are coprime, that exponent is a multiple of $mn$. This being true for every $p$, using the lemma again tells us that $k$ is a perfect $mn$-th power.