[Math] Prove that if $n$ and $m$ are positive integers such that $n^n|m^m$ then $n|m$.

Question

This is how I did it, but not sure if it is a correct proof.

Assume that $n^n | m^m$. And we write $m= p_1^{a_1}p_2^{a_2}…p_k^{a_k}$ and $n=q_1^{b_1}q_2^{b_2}…q_l^{b_l}$.
So,
$$m^m = (p_1^{a_1}p_2^{a_2}…p_k^{a_k})^m = c \ . \ (q_1^{b_1}q_2^{b_2}…q_l^{q_l})^n$$ for some positive integer $c$ which requires that for any $q_i$ there exist some $p_j$ such that $q_i=p_j$ and of course $ nb_i \le ma_j$. But this doesn't not imply $b_i \le a_j$ meaning I failed to prove $n|m$.

On the other hand whatever example I try I can't come to a counterexample.

Is it true if $n^n|m^m$ then $n|m$ for $n$ and $m$ are positive integers?

Answer 1

$4^4|10^{10}$ but $4 \not\mid\ 10$.