I am trying to prove this lemma in group theory.
For $a \in G$, if $a^m = e$ for some $m \in \mathbb{N}$, then the order of $a$ must divide $m$.
Proof. For a contradiction, suppose the order of $a$, denoted $b$, does not divide $m$. By the Euclidean algorithm, we may write $$m = nb + r$$ for $n \in \mathbb{N}$ and $0 < r < b$. That $r > 0$ follows from the assumption toward a contradiction. (Otherwise, it could be the case that $b \mid m$.) We have: \begin{align*} a^m = a^{nb + r} = a^{nb} a^r = a^{nb} a^r = (a^b)^n a^r = e^n a^r = a^r. \end{align*} Since $r < b$, and $b$ is the smallest positive integer such that $a^b = e$, $a^r \neq e$, so $a^m \neq e$, a contradiction. Hence, we have $r = 0$, so $b \mid m$.
This seems correct to me. Generally, proofs without contradiction are seen as cleaner, so you might want to do the following: assume $m=nb+r$ with $0\leq r<b$, and then derive $a^r=1$, which directly implies $r=0$. However, your proof is certainly correct.