 Solution

For an $m\times n$ rectangle, let $S_j$ be the symbols absent from column $j$. Each has size $n-m$. Across all columns, each symbol is absent exactly $n-m$ times, because it appears once in each existing row. For any set $J$ of columns, count incidences between those columns and their missing symbols. There are $|J|(n-m)$ incidences, while each distinct symbol in the union contributes at most $n-m$. Therefore $|{\bigcup }_{j\in J}{S}_j|\ge |J|$.

Hall's theorem supplies distinct representatives, one per column, forming an additional row with all $n$ symbols. This extends the rectangle without any column repetition. Repeat with $m+1$ rows, then successively until there are $n$ rows. Hence every finite latin rectangle extends to a Latin square by adding the required rows.