Proposition
Let $A\in\mathfrak{M}_{(m+n)\times(m+n)}(\mathbb{K})$, $B\in\mathfrak{M}_{n}(\mathbb{K})$, $D\in\mathfrak{M}_{m}(\mathbb{K})$. If $A=\left(\begin{array}{cc}B& C\\0 & D \end{array}\right)$ and $B$ is invertible, then $\det A = \det B \cdot \det D$.
My proof Since $B$ is invertible, applying row transformations $ \underbrace{\left(\begin{array}{cc}E_k & 0 \\ 0 & I\end{array}\right) \cdots \left(\begin{array}{cc}E_1 & 0 \\ 0 & I\end{array}\right)}_{Y} \left(\begin{array}{cc}B & C \\ 0 & D\end{array}\right) = \left(\begin{array}{cc}H_B & C' \\ 0 & D\end{array}\right) = \left(\begin{array}{cc}I_n & C' \\ 0 & D\end{array}\right)$ where $H_B$ is the Hermitian form of $B$., and $\det Y = \det(E_k\cdots E_1) = \det B^{-1} = (\det B)^{-1} $.
Now, applying row transformations again: $ \underbrace{\left(\begin{array}{cc}F_l & 0 \\ 0 & I\end{array}\right) \cdots \left(\begin{array}{cc}F_1 & 0 \\ 0 & I\end{array}\right)}_{Z} \left(\begin{array}{cc}H_B & C' \\ 0 & D\end{array}\right) = \underbrace{\left(\begin{array}{cc}H_B & C' \\ 0 & H_D\end{array}\right)}_{W} $ let suppose that that $D$ is invertible, then $H_D = I_m$ and, since $W$ is triangular, $\det W = 1$, $\det Z = \det(F_l\cdots F_1) = (\det D)^{-1}$ and $\det A = \det B \cdot \det D$.
If $D$ is not invertible, $\det D = 0$, $\det A = 0$ since there exist null rows and $\det A = \det B \cdot \det D$.
Is there anything incomplete/wrong? Is there a faster/more elegant proof? Thanks in advance.