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 $\det(\mathbf{M}) = (-1)^k \det(\mathbf{A}) \det(\mathbf{B})$?

I think julien has answered your question in a beautiful way. Here I will put your question in a slightly more general context. Note that for a partitioned matrix
$$
M=\pmatrix{P&Q\\ R&S},
$$
where the four submatrices are square and have the same size, if $R$ commutes with $S$ (i.e. $RS=SR$), we have the block determinant formula
$$\det M=\det(PS-QR)\tag{1}$$
that is analogous to the determinant formula for $2\times2$ matrices. So, if we put $P=S=0,\ Q=A$ and $R=B$, we get $\det M=\det(-AB)=(-1)^k\det(AB)=(-1)^k\det(A)\det(B)$.
Certainly you shouldn't do your exercise using this formula, because an exercise like yours are often intended to be solved in a more elementary way, and $(1)$ is more difficult to prove than your problem statement. Nonetheless it's good to know some nice formulae. Yet formula $(1)$ should be applied with care: in general, we have
$$
\det M=
\begin{cases}
\det(PS-QR) & \text{ if } RS=SR,\\
\det(SP-RQ) & \text{ if } PQ=QP,\\
\det(SP-QR) & \text{ if } QS=SQ,\\
\det(PS-RQ) & \text{ if } PR=RP.
\end{cases}\tag{2}
$$
Since the four determinants $\det(PS-QR),\,\det(SP-RQ),\,\det(SP-QR)$ and $\det(PS-RQ)$ are in general different, the pair of submatrices that commute as well as the order of the four blocks do matter.
