Matrix Multiplication is Associative

Showing $\left( A B \right) C = A \left( B C \right)$ is relatively a simple matter using the super duper summation convention.
Firstly the i^th row, j^th column of a matrix $A$ is designated by $A_{ij}$, $A^{i}_{j}$ or $A^{ij}$. Matrix multiplication is then defined as $$\left[ XY \right]_{ij} = X_{ik}Y_{kj}$$ where square brackets have been placed around $X Y$ to designate the $i j$ element of the matrix $X \cdot Y$. $$\begin{aligned} \left[ \left( A B \right) C \right]_{ij} & = \left( A B \right)_{ik} C_{kj} \\ & = A_{il} B_{lk} C_{kj} \\ & = A_{il} \left( B_{lk} C_{kj} \right) \\ & = A_{il} \left[ B C \right]_{lj} \\ & = \left[ A \left( B C \right) \right]_{ij} \\ \therefore \left( A B \right) C & = A \left( B C \right) \end{aligned}$$

Matrix Inverses are unique

Left Inverse = Right Inverse

Let $A$ be an $n \times n$ matrix and $AB = \mathbb{I}$, so that $B = A^{-1}$.
It follows that $BA = \mathbb{I}$...
$$\begin{aligned} AB & = \mathbb{I} \\ B \left( AB \right) & = B \mathbb{I} \\ \left( B A \right) B & = B \\ \therefore \left( B A \right) & = \mathbb{I} \end{aligned}$$

Inverses Are Unique

Suppose another inverse of $A$ existed, $C$, so that $AC = CA = \mathbb{I}$.
It follows that $B = C = A^{-1}$...
$$\begin{aligned} A B & = \mathbb{I} \\ C \left( A B \right) & = C \mathbb{I} \\ \left( C A \right) B & = C \\ \left( \mathbb{I} \right) B & = C \\ \therefore B & = C \end{aligned}$$