Prove that multiplication of residue classes in $\mathbb{Z}/n\mathbb{Z}$ is associative.
Let $0\le a,b,c < n$. By associativity of integers and the division algorithm, \begin{align*} (ab)c &= qn+r = a(bc) & \text{for some integer }q \ge 0 \text{ and } 0\le r < n \end{align*}
Let $ab = nq_1 + r_1, bc=nq_2+r_2$ for some integers $q_1,q_2\ge 0$ and $0\le r1,r2 < n$. \begin{align*} (nq_1 + r_1)c &= a(nq_2 + r_2) \\ nq_1c + r_1c &= anq_2 + ar_2 \end{align*}
Let $r_1c = nq_3 + r_3, ar_2=nq_4 + r_4$ for some integers $q_3,q_4\ge 0$ and $0\le r_3,r_4 < n$. \begin{align*} nq_1c + (nq_3 + r_3) & = anq_2 + (nq_4 + r_4) \\ n(q_1c + q_3) + r_3 &= n(aq_2+q_4) + r_4 \end{align*}
This implies $q=q_1c+q_3=aq_2+q_4$ and $r_3=r_4$.
Note
\begin{align*} ab \equiv r_1 &\text{ and } bc\equiv r_2 (\operatorname{mod} n) \\ r_1 c \equiv r_3 &\text{ and } ar_2\equiv r_4 (\operatorname{mod} n) \\ \therefore (\bar{a}\bar{b}) \bar{c} =\bar{r_3} &= \bar{r_4} = \bar{a} (\bar{b}\bar{c}) \end{align*}
Is this a reasonable argument?