To prove that $(A B) C=A(B C)$, use the column vectors $b_{1}, \ldots . b_{n}$ of $B$. First suppose that $C$ has only one column $c$ with entries $c_{1}, \ldots, c_{n}$ :
$A B$ has columns $A b_{1}, \ldots, A b_{n}$ and $B c$ has one column $c_{1} b_{1}+\cdots+c_{n} b_{n}$.
Then $(A B) c=c_{1} A b_{1}+\cdots+c_{n} A b_{n}$ equals $A\left(c_{1} b_{1}+\cdots+c_{n} b_{n}\right)=A(B c)$.
Linearity gives equality of those two sums, and $(A B) c=A(B c)$. The same is true for all other _ of $C$. Therefore $(A B) C=A(B C)$.