Hint $\ $ It's a consequence of Sylvester's determinant identity $\rm\:det(1 + AB) = det(1+BA),\:$ which has a very simple universal proof: $ $ over the polynomial ring $\rm\ \mathbb Z[A_{\,i\,j},B_{\,i\,j}\,]\ $ take the determinant of $\rm\, (1+AB)\, A = A\, (1+BA)\ $ then cancel $\rm\, det(A)\ $ (valid since the ring is a domain). $ $ Extend to non-square matrices by padding appropriately with $0$'s and $1$'s to get square matrices. Note that the proof is purely algebraic - it does not require any topological notions (e.g. density).
Alternatively $\ $ Proceed by way of Schur decomposition, namely
$\rm\left[ \begin{array}{ccc} 1 & \rm A \\ \rm B & 1 \end{array} \right]\ =\ \left[ \begin{array}{ccc} 1 & \rm 0 \\ \rm B & 1 \end{array} \right]\ \left[ \begin{array}{ccc} 1 & \rm 0 \\ \rm 0 & \rm 1-BA \end{array} \right]\ \left[ \begin{array}{ccc} 1 & \rm A \\ \rm 0 & 1 \end{array} \right]$
$\rm\phantom{\left[ \begin{array}{ccc} 1 & \rm B \\ \rm A & 1 \end{array} \right]}\ =\ \left[ \begin{array}{ccc} 1 & \rm A \\ \rm 0 & 1 \end{array} \right]\ \left[ \begin{array}{ccc} \rm 1-AB & \rm 0 \\ \rm 0 & \rm 1 \end{array} \right]\ \left[ \begin{array}{ccc} 1 & \rm 0 \\ \rm B & 1 \end{array} \right]$
See my posts in this sci.math thread on 09 Nov 2007 for further discussion.