How to prove that $\det(M) = (-1)^k \det(A) \det(B)?$ Let $\mathbf{A}$ and $\mathbf{B}$ be $k \times k$ matrices and $\mathbf{M}$ is the block matrix
$$\mathbf{M} = \begin{pmatrix}0 & \mathbf{B} \\ \mathbf{A} & 0\end{pmatrix}.$$
How to prove that $\det(\mathbf{M}) = (-1)^k \det(\mathbf{A}) \det(\mathbf{B})$?

Here is one way among others:
$$
\left( \matrix{0&B\\A&0}\right)=\left( \matrix{0&I_k\\I_k&0}\right)\left( \matrix{A&0\\0&B}\right).
$$
I assume you are allowed to use the block diagonal case, which gives $\det A\cdot\det B$ for the matrix on the far right. Just in case, this follows for instance from Leibniz formula. 
Now it boils down to 
$$
\det\left( \matrix{0&I_k\\I_k&0}\right)=(-1)^k.
$$
This is the matrix of a permutation made of $k$ transpositions. So the determinant is $(-1)^k$ after $k$ elementary operations (transpositions  in this case) leading to the identity matrix.
