I really like proofs that neatly use block matrices. For example, Sylvester's determinant identity has a nice proofs along these lines such as these, as do the many formulas which use Schur complements. I was wondering if a proof along these lines would work for Cauchy-Binet.
The formula can be expressed as follows:
If $A$ is $m \times n$ and $B$ is $n \times m$ (where $m \leq n$), we have $$ \det(AB) = \sum_{S\in\tbinom{[n]}m} \det(A_{[m],S})\det(B_{S,[m]}) $$
Where $\tbinom{[n]}m$ is the collection of size $m$ subsets of $[n] = \{1,\dots,n\}$.
Ideally, I'd like to find block-matrices $M,N$ such that $\det(MN)$ is clearly $\det(AB)$, and $\det(NM)$ gives us the other side.
My instinct is to just "pad our matrices out" into square matrices, but look at $M = \pmatrix{A\\0}$ and $N = \pmatrix{B & 0}$ clearly fails. Perhaps there's something nice with $M = \pmatrix{0&A\\A^T&0}$.
