= Solution
Over a commutative field, define the <determinant> by the signed permutation sum
$$
\boxed{\det A=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^n a_{i,\sigma(i)}}.
$$
It is linear in each row separately. Swapping two rows reindexes the permutations by a transposition and reverses the sign; thus a <matrix> with two equal rows has <determinant> zero. Adding $c$ times row $i$ to a different row $j$ consequently gives
$$
\det A'=\det A+c\det(\text{matrix with row }j\text{ replaced by row }i)=\det A.
$$
These are immediate consequences of the definition, not assumptions about elementary row operations.
For the block upper-triangular <matrix>, a nonzero permutation term must assign every bottom row to one of the last $n$ columns, because its first $n$ entries are zero. Those rows use all the last columns; the top rows must use all the first columns. The sign and products factor into two independent $n$-permutations, giving
$$
\boxed{\det\begin{pmatrix}A&B\\0&C\end{pmatrix}=\det A\det C}.
$$
In particular $X_0=\begin{pmatrix}B&I\\0&A\end{pmatrix}$ has <determinant> $\det B\det A$. For each bottom row $n+i$, subtract $a_{ij}$ times top row $j$ for $j=1,\ldots,n$. The top rows remain unchanged, so these additions produce
$$
X_1=\begin{pmatrix}B&I\\-AB&0\end{pmatrix}
$$
without changing the <determinant>. Exchange its first and last blocks of $n$ columns; the permutation has sign $(-1)^{n^2}$ and gives $\begin{pmatrix}I&B\\0&-AB\end{pmatrix}$. Hence
$$
(-1)^{n^2}\det X_1=\det(-AB)=(-1)^n\det(AB).
$$
Since $n^2$ and $n$ have the same parity, $\det X_1=\det(AB)$. Combining the two evaluations proves
$$
\boxed{\det(AB)=\det A\det B}.
$$
No invertibility of $A$ or $B$ was used, so the proof includes singular <matrices>.
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