$n|m \Leftrightarrow m\mathbb{Z} \le n\mathbb{Z}$ [closed]

How can we prove that: $n|m \Leftrightarrow m\mathbb{Z} \le n\mathbb{Z}$
I understand that if $m\mathbb{Z} \le n\mathbb{Z}$ so that $m=m\cdot 1\in n\mathbb{Z}$ and then $n|m$ but I don't understand why it's true.

Let $G $ be a group and $A,B \leq G$. If $A \subseteq B$, then $A \leq B$ (Try to prove it if you haven't already). So in this case you just need to prove that $m \mathbb Z \subseteq n\mathbb Z$.

I'll mark the proof as a spoiler just in case you want to try it yourself first:

Let $x \in m\mathbb Z$. Then $x = mh$, for some $h \in \mathbb Z$. But $n|m$, meaning that $m = nq$, for some $q \in \mathbb Z$, so we can conclude that $x = mh = n \cdot qh \in n\mathbb Z$. So $m\mathbb Z\subseteq n \mathbb Z$, emaning that $m \mathbb Z \leq n\mathbb Z$