A question about the matrix tree theorem. How to compute the determinant of the following matrix?

The problem is to determine the number of spanning trees in $$K_{r,s}$$. The Matrix tree theorem states that the answer is the determinant of the matrix $$M_{ii}$$, where $$M_{ii}$$ is the matrix obtained from deleting the first row and the first column of $$X-A$$, where $$X$$ is the diagonal matrix of degree of each vertex and $$A$$ is the adjacency matrix of $$K_{r,s}$$.

I have the desired matrix below, denoted as $$M$$, and its determinant is the answer of the graph theory question. However, I have difficulty in computing its determinant. The upper left corner is an $$(r-1) \times (r-1)$$ diagonal matrix with diagonal entry $$s$$, and the lower right corner is an $$s \times s$$ diagonal matrix with diagonal entry $$r$$. All the other entries are $$-1$$.

Though the matrix looks rather harmonic, I have no idea how to compute its determinant when $$r$$ and $$s$$ are arbitrary natural numbers. I appreciate any help from you guys.

Here is a convenient trick for computing the determiant: instead of using any expansion formulas, find a full set of eigenvalues or eigenvectors for $$M$$. Then, use the fact that the determinant is the product of eigenvalues. This is often a good strategy when the matrix has "nice" symmetries, and the symmetries can be used to find all eigenvalues easily. Also, this is a good strategy when a large matrix has only a small number of (distinct) eigenvalues, which is the case here!
Now consider the Laplacian matrix of the graph $$K_{r,s}$$ (before deleting a row and column): the eigenvalues are $$0$$, $$r$$, $$s$$, and $$r+s$$. The $$r$$-eigenspace has multiplicity $$s-1$$, the $$s$$-eigenspace has multiplicity $$r-1$$, and the $$(r+s)$$-eigenspace and $$0$$-eigenspace have multiplicity $$1$$. Therefore the determinant of the reduced Laplacian (after deleting a row and column) is $$\frac1{n} \prod (\text{non-zero eigenvalues}) = \frac1{r+s}r^{s-1}s^{r-1} (r+s) = r^{s-1} s^{r-1}.$$ (Once we know the eigenvalues are $$r$$, $$s$$, and $$r+s$$, it is straightforward to check their multiplicities. The part that requires some luck / insight is to guess that these are the correct eigenvalues.)
You can grind it out using row and column operations, to end up with an upper diagonal matrix, from which the determinant is the product of the diagonals, or a quicker way is to use the formula \begin{align} \text{det} \begin{pmatrix} A & B \\ C & D \end{pmatrix} = \text{det}(A) \text{det}(D - CA^{-1}B) \end{align}
which is not so hard to compute, as A is diagonal, so the inverse has a nice form. I think the answer should be $$r^{s-1}s^{r-1}$$.