I'm trying to prove that matrix multiplication is associative, but seem to be making mistakes in each of my past write-ups, so hopefully someone can check over my work.
Theorem. Let $A$ be $\alpha \times \beta$, $B$ be $\beta \times \gamma$, and $C$ be $\gamma \times \delta$. Prove that $(AB)C = A(BC)$.
Proof. Define general entries of the matrices $A$, $B$, and $C$ by $a_{g,h}$, $b_{i,j}$, and $c_{k,m}$, respectively. Then, for the LHS: \begin{align*} & (AB)_{\alpha, \gamma} = \sum\limits_{p=1}^{\beta} a_{\alpha,p} b_{p,\gamma} \\ & \left((AB)C\right)_{\alpha, \delta} = \sum\limits_{n=1}^{\gamma} \left(AB\right)_{\alpha, n} c_{n, \delta} = \sum\limits_{n=1}^{\gamma} \left(\sum\limits_{p=1}^{\beta} a_{\alpha,p} b_{p,n} \right) c_{n, \delta} = \sum\limits_{n=1}^{\gamma} \sum\limits_{p=1}^{\beta} \left(a_{\alpha,p} b_{p,n}\right) c_{n, \delta}. \end{align*} For the RHS: \begin{align*} & \left(BC\right)_{\beta, \delta} = \sum\limits_{n=1}^{\gamma} b_{\beta, n} c_{n, \delta} \\ & \left(A\left(BC\right)\right)_{\alpha,\delta} = \sum\limits_{p=1}^{\beta} a_{\alpha,p} (BC)_{p, \delta} = \sum\limits_{p=1}^{\beta} a_{\alpha,p} \left(\sum\limits_{n=1}^{\gamma} b_{p, n} c_{n, \delta} \right) = \sum\limits_{p=1}^{\beta} \sum\limits_{n=1}^{\gamma} a_{\alpha,p} \left(b_{p, n} c_{n, \delta} \right). \end{align*} Assuming I have written these correctly, we can make two observations: first, the summands are equivalent, as multiplication is associative. Second, the order of the summations doesn't matter when we're summing a finite number of entries. Thus, $(AB)C = A(BC)$.
How does this look?
Your proof is fine.
We can change the order of summation as the sum is finite. When we mention multiplication is associative, we might want to mention multiplicative of which object, such as multiplicative of real numbers or complex number.