Let $A\in\Ma(m+n,\R)$ be the block matrix
$$
A=\left(
\begin{array}{cc}
A_{11}&A_{12}\\
A_{21}&A_{22}
\end{array}
\right)~.
$$
If $A_{11}\in\Ma(m,\R)$ is invertible then
$$
\det A=\det(A_{11})\det(A_{22}-A_{21}A_{11}^{-1}A_{12})~.
$$
We have
$$
\left(\begin{array}{cc}
1&0\\
-A_{21}A_{11}^{-1}&1
\end{array}\right)
\left(\begin{array}{cc}
A_{11}&A_{12}\\
A_{21}&A_{22}
\end{array}\right)
\left(\begin{array}{cc}
1&-A_{11}^{-1}A_{12}\\
0&1
\end{array}\right)
=\left(\begin{array}{cc}
A_{11}&0\\
0&A_{22}-A_{21}A_{11}^{-1}A_{12}
\end{array}\right)~.
$$
Since the determinants of the first and the third matrix are $1$ (cf. exam), we conclude by exam that
$$
\det(A)=\det(A_{11})\det(A_{22}-A_{21}A_{11}^{-1}A_{12})~.
$$