For positive integers $m$ and $n$, if $10m\mid 21n$, then $10\mid n$

Question

Let $m$ and $n$ be positive integers. If $10m\mid 21n$, then $10\mid n$.

This is what I've done so far:

If $10m\mid 21n$ then $21n = (10m)k$.

Since $m$ and $k$ are just integers, let $m\cdot k= p$. Therefore, $21n=10p$ so $10\mid 21n$.

Since $GCD(10,21)=1$, by Euclid's lemma $10\mid n$.

Any advice? Do you think my proof is correct?

Thanks!