Sub-determinants of an orthogonal matrix

Let $A$ be a matrix in the special orthogonal group, $A \in \mathrm{SO}_n$. This means that $A$ is real, $n \times n$, $A^t A = I$ and $\det(A)=1$, that is, the column vectors of $A$ make a positively-oriented orthonormal basis for $\mathbb R^n$.

Decompose $A$ as a block matrix

$$A = \begin{pmatrix} B & C \\ D & E\end{pmatrix}$$

where $B$ is $k \times k$ and $E$ is $(n-k)\times (n-k)$.

I'm looking for a basic linear-algebra argument that $\det(B) = \det(E)$, ideally something that could be presented in a 2nd or 3rd year undergraduate course. So I do not want people to invoke anything like tensor products or differential forms.

Some googling brought up this, Hudson's book "Kummer's quartic surface", containing the following argument.

Given $A = \begin{pmatrix} B & C\\ D & E\end{pmatrix}$, orthgonality gives us that $A^{-1} = A^t = \begin{pmatrix} B^t & D^t\\C^t & E^t\end{pmatrix}$.

The block matrix multiplication gives $$I = A^t A = \begin{pmatrix} B^tB + D^tD & B^t C + D^tE\\C^tB + E^t D & C^tC + E^tE\end{pmatrix}$$

so $B^tB + D^tD = I$, $B^tC + D^tE = 0$, and $C^tC + E^tE = I$ (but the two $I$s probably have different size).

Now, using these relations, it's easy to see that $$\begin{pmatrix} B^t & D^t\\0&I\end{pmatrix} A = \begin{pmatrix} B^t & D^t\\0&I\end{pmatrix}\begin{pmatrix}B&C\\D&E\end{pmatrix} = \begin{pmatrix} I & 0\\D &E\end{pmatrix}$$

(but I confess to being at a loss of how to motivate where this came from - it's in the book).

From here, taking the determinant of both sides, using the fact that $\det(A) = 1$ and that $\det$ is a homomorphism, gives $$\det\begin{pmatrix} B^t & D^t\\0&I\end{pmatrix} = \det\begin{pmatrix} I&0\\D&E\end{pmatrix}$$

but the determinant of a block matrix with one block $0$ is the product of the determinants of the diagonal blocks, so we get $\det(B^t) = \det(E)$. Finally, note that $\det(B^t) = \det(B)$, so we're done.