Block matrix determinant formula

This is more general result $$\det\pmatrix{A&B\\0&D} = \det A\det D$$ and to prove it notice that

$$\pmatrix{A&B\\0&D}=\pmatrix{I_k&0\\0&D}\pmatrix{A&B\\0&I_n}$$ and we develop along the $k$ first rows we find $$\det\pmatrix{I_k&0\\0&D}=\det D$$ and along the last $n$ rows we find $$\det\pmatrix{A&B\\0&I_n}=\det A$$ and the result follows.

If your matrices have coefficients in an integral domain, you can pass to the field of fractions and take an algebraic closure to use the Jordan canonical form, in which case this equation becomes trivial since both sides of the equation are the product of the products of the eigenvalues of $A$ and $D$.

Otherwise you can use Laplace expansion and notice that the sum goes over all the permutations of $S_{k+n}$, in which case the only terms which "survive" are the permutations which map $\{ 1,\cdots, k\}$ to itself and $\{k+1,\cdots,k+n\}$ to itself. Then you can split the sum in two parts and obtain $\det(A) \det(D)$. This proof works over any (commutative) ring.

Hope that helps,