Does $a^n \mid b^n$ imply $a\mid b$?

If you can assume the fundamental theorem of arithmetic (that each integer has a unique factorization in prime numbers), you can write: $$ \begin{align*} a &= p_1^{e_1} p_2^{e_2} \ldots p_r^{e_r} \\ b &= q_1^{d_1} q_2^{d_2} \ldots q_s^{d_s} \end{align*} $$ Here the $p_i$, $q_i$ are primes, and $e_i$ and $d_i$ are all greater than 0. If $a^n \mid b^n$, then $p_i^{n e_i}$ must have a counterpart in a $q_j^{n d_j}$, in that $p_i = q_j$ and $n e_i \le n d_j$, so it must then also be that $e_i \le d_j$; and this means $a \mid b$.

Hint $\ $ Either examine exponents in unique prime factorizations, or, by the Rational Root Test, the reduced rational root $\rm\:x = b/a\:$ of $\rm\:x^n = c\in\Bbb Z\:$ must be integral, so $\rm\:b/a\in\Bbb Z\:\Rightarrow\:a\mid b.$

Hint: $p$ is a prime factor of $k$ if and only if $p^n$ is a factor of $k^n$. This holds for any prime $p$, integer $k$, and positive integer $n$.