Let $a_1...a_k$ be positive integers that are pairwise relatively prime. Assume that $\sqrt[m]{a_1...a_k}\in\mathbb{N}$ for some $m\in\mathbb{N}$. Show that $\sqrt[m]{a_i}\in\mathbb{N}$ for each $i$.
I want to know if this proof is correct. I have trouble with constructing rigorous proofs.
So I know that for root m to be an element of $\mathbb{N}$, $a_1...a_k$ must have m copies of any natural number, say $s\in\mathbb{N}$. However, since each $a_i$ must be pairwise relatively prime, each $a_i$ cannot share any common factors, this includes s. But if any $a_i$ does not have a factor of s, then it cannot be taken out of the root which would produce a product $a_1...a_k\notin\mathbb{N}$. So each $a_i$ must be reducible by root m.