= Solution
For the specified row and column, expand the two <matrix products> and reorder only <finite sums>:
$$
\begin{aligned}
((AB)C)_{ij}
&=\sum_{k=1}^{p}\left(\sum_{\ell=1}^{n}A_{i\ell}B_{\ell k}\right)C_{kj}\\
&=\sum_{\ell=1}^{n}A_{i\ell}\left(\sum_{k=1}^{p}B_{\ell k}C_{kj}\right)
=(A(BC))_{ij}.
\end{aligned}
$$
The middle equality uses <scalar> <associativity> and <distributivity>, and the interchange of two <finite sums>. Together with their equal <dimensions>, equality of every entry proves \b[$(AB)C=A(BC)$].
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