How can I show that $a^n|b^n \Rightarrow a|b$ How can I show the following
  $$a^n|b^n \Rightarrow a|b$$ with $a,b$ integers.


$$a^n|b^n \Rightarrow b^n=m \cdot a^n \Rightarrow b^n=(m\cdot a^{n-1}) \cdot a\qquad(1)$$
How can I continue? Do I maybe have to suppose the opposite and arrive at contradiction?
$$\text{So } a \nmid b \Rightarrow b=q\cdot a+r$$
Replacing this at the relation $(1)$ could I conlude to something to get a contradiction?
Or is there an other way to prove this??
Hint $ $ Clear if $\,a=0.$ Else $\,(\color{#c00}{b/a})^n \! = k\in\Bbb Z\,\Rightarrow\, b/a\in \Bbb Z,\,$ because the $ $ Rational Root Test $ $ implies that $ $ any $ $ rational $ $ root $ $ of $\ \ \color{#c00}x^n - k\ $ is integral.
Assuming, for contradiction, that $a \nmid b$, there must be some prime $p$ to some power $m$ such that $p^m | a$, but $p^m \nmid b$. Then $p^{nm} | a^n$, but $p^{nm}\nmid b$ by simple counting of prime factors.