Show determinant of $\left[\begin{matrix} A & 0 \\ C & D\end{matrix}\right] = \det{A}\cdot \det{D}$ Let $A \in \mathbb{R}^{n, n}$, $B \in \mathbb{R}^{n, m}$, $C \in \mathbb{R}^{m, n}$ and $D \in \mathbb{R}^{m, m}$ be matrices.
Now, I have seen on Wikipedia the explanation of why determinant of $\left[\begin{matrix} A & 0 \\ C & D\end{matrix}\right] = \det{A}\cdot \det{D}$, but I still did not get it.
Specifically, the explanation is:

This can be seen ... from a decomposition like:
$\left[\begin{matrix} A & 0 \\ C & D\end{matrix}\right] = \left[\begin{matrix} A & 0 \\ C & I_{m}\end{matrix}\right]\left[\begin{matrix} I_n & 0 \\ 0 & D\end{matrix}\right]$

I understood that the equation is true from the standard rules of matrix-matrix multiplication, but it is still not too clear why this should prove what we want to prove or show.
If $A$, $B$, $C$ and $D$ were regular reals (and $I_{i}$ was $1$), then the equation and the explanation would be obvious, because of the standard rules of calculating determinants...
But in this case, I cannot understand why the equation shows that the final determinant is $$\det{A} \cdot \det{D}$$
Those 2 matrices $\left[\begin{matrix} A & 0 \\ C & I_{m}\end{matrix}\right]$ and $\left[\begin{matrix} I_n & 0 \\ 0 & D\end{matrix}\right]$ basically could not be triangular or diagonal matrices, from my understanding...

The first thing to note is of course the fact that $\det (AB)=\det A \cdot\det B$. This is well known - if you search you will find several proofs - in some texts this condition is used as an axiom when defining the determinant. So this allows you to assert 
\begin{equation} \det \begin{bmatrix} A & 0 \\ C & D\end{bmatrix} = \det \begin{bmatrix} A & 0 \\ C & I_{m}\end{bmatrix} \det \begin{bmatrix} I_n & 0 \\ 0 & D\end{bmatrix}.\end{equation}
Now it is very simple - if you just apply Laplace expansion (cofactor expansion - which is the usual way to determine the determinant by hand) the answer follows directly, since you can expand 
\begin{equation}\det \begin{bmatrix} A & 0 \\ C & I_{m}\end{bmatrix}\end{equation} starting at the last column and then the second last column etc. to see that 
\begin{equation}\det \begin{bmatrix} A & 0 \\ C & I_{m}\end{bmatrix}=1^m \cdot \det A = \det A.\end{equation} Now similarly you can expand 
\begin{equation} \det \begin{bmatrix} I_n & 0 \\ 0 & D\end{bmatrix} \end{equation} by the first row, then the second, etc. to see that 
\begin{equation} \det \begin{bmatrix} I_n & 0 \\ 0 & D\end{bmatrix}=1^n \cdot \det D = \det D.\end{equation}
If you have difficulty understanding this last part (cofactor expansion), then I think you should spend some time with a good text book to make sure you understand exactly how it works. Basically it allows you to calculate determinants very easily in some cases by making intelligent choices with regards to the rows and columns you are expanding by. In this case in particular there are rows/columns which contain only a single 1 which effectively means you can reduce immediately to determining the cofactor, or minor for that matter since the 1 is in a diagonal position.
