Proof that if $a^n|b^n$ then $a|b$ I can't get to get a good proof of this, any help?
What I thought was:
$$b^n = a^nk$$
then, by the Fundamental theorem of arithmetic, decompose $b$ such:
$$b=p_1^{q_1}p_2^{q_2}...p_m^{q_m}$$
with $p_1...p_m$ primes and $q_1...q_n$ integers.
then 
$$b^n=(p_1^{q_1}p_2^{q_2}...p_m^{q_m})^n= p_1^{q_1n}p_2^{q_2n}...p_m^{q_mn}$$
but here i get stucked, and i can't seem to find a good satisfactory way to associate $a$ and $b$...
Any help will be appreciated 

Hint $\ p^{n\alpha}\!\mid p^{n\beta}\! \iff n \alpha \le n\beta \iff \alpha \le \beta \iff p^\alpha\mid p^\beta.\,$ Apply it to prime factorizations of a,b.
Simpler: $\, (b/a)^n = k\in\Bbb Z\Rightarrow b/a\in \Bbb Z\,$ by the Rational Root Test applied to $\,x^n - k.$
