The aim of the exercise is to prove that $$\det\begin{pmatrix}B&D\\0&C\end{pmatrix}=\det(B)\det(C)$$ where $B\in \mathbb R^{p\times p}$, $C\in \mathbb R^{q\times q}$, $n=p+q$ and $M:=\begin{pmatrix}B&D\\0&C\end{pmatrix}\in\mathbb R^{n\times n}$.
Let $\mathcal B_n(e_1,...,e_n)$ the canonical basis of $\mathbb R^n$, $E=Span\{e_1,...,e_p\}$ and $E'=Span\{e_{p+1},...,e_n\}$. Therefore $\mathbb R^n=E\oplus E'$.
Set $M=(m_{j})_{1\leq j\leq n}$ where $m_{j}$ are the column of $M$ and are defined as $m_j=b_j+0$ if $1\leq j\leq p$ and $m_j=d_j+c_j$ if $p+1\leq j\leq n$. Therefore
$$\det(M)=\det_{\mathcal B_n}(m_1,...,m_n)=\det_{\mathcal B_n}(b_1,...,b_p,c_{p+1}+d_{p+1},...,c_n+d_n).$$
Let $$\alpha :E^p\to \mathbb R$$ defined as $$\alpha (x_1,...,x_p)=\det_{\mathcal B_n}(x_1,...,x_p,c_{p+1}+d_{p+1},...,c_n+d_n).$$ Recall that $\{e_1,...,e_p\}$ is a basis of $E$. Therefore, by linearity : $$\alpha (e_1,...,e_p)=\det_{\mathcal B_n}(e_1,...,e_p,c_{p+1},...,c_n+d_n)+\underbrace{\det_{\mathcal B_n}(e_1,...,e_p,d_{p+1},...,c_n+d_n)}_{=0},$$ but I don't understand why the last term is $0$.
Here is another way:
Note that if $C$ is invertible we have $\begin{bmatrix} B & D \\ 0 & C\end{bmatrix} = \begin{bmatrix} I & DC^{-1} \\ 0 & I\end{bmatrix} \begin{bmatrix} B & 0 \\ 0 & C\end{bmatrix}$ and so we see that $\det \begin{bmatrix} B & D \\ 0 & C\end{bmatrix} = = \det B \det C$.
Since this is true for all invertible $C$, $\det$ is continuous and the invertible matrices are dense we see that it is true for all $C$.