(Must use induction) proving multiplication of determinant 
Prove that if $M \in M_{n,n}(F)$ can be written in the form
  $M=\begin{pmatrix} A & B\\ O & C \end{pmatrix}$, where $A$ and $C$ are square matrices, then $\det(M)=\det(A)\det(C)$

I know this question has many fancy answers, but they are using more advanced techniques. I am told to prove by induction using cofactor expansion.
here is my current work:
We proceed by induction on $M$
Let $M$ be a $2$ by $2$ matrix, then it clearly holds.
Assume $n-1$ by $n-1$ matrix holds for $M$, and we try to prove $n$ by $n$ matrix for $M$. Let $A$ be a $n-m$ by $n-m$ matrix and $C$ a $m$ by $m$ matrix.
From the definition of cofactor expansion of determinant, $$\det(M)=(-1)^{1+1}*a_{11}*\det(\hat A_{11})+...+(-1)^n*a_{1n}*\det(\hat A_{1n})$$
Since we are expanding along the first row, and it holds true for $\det(\hat A_{ij})$ ( $n-1$ by $n-1$ matrix), we rewrite as $$=\det(C)*((-1)^{2}*a_{11}*\det(A^{(1,1)})+...+(-1)^n*a_{1n}*\det(A^{1,n})=\det(C)*\det(A),$$ where $A^{(i,j)}$ indicates the position of row and column cut by expansion.
The induction hypothesis applies when we move to the second row, third row... Hence proved.
Can anyone tell if I am on the right track for this?
 A: You were on the right track with your other question, which was close as being a duplicate.
Since we can write $\begin{pmatrix} A & B \\ 0 & C \end{pmatrix}$ as  $\begin{pmatrix} I & 0 \\ 0 & C  \end{pmatrix} \cdot \begin{pmatrix} A & B \\ 0 & I \end{pmatrix}$
We will prove by induction that 
$$\det\begin{pmatrix} I & 0 \\ 0 & C  \end{pmatrix}=\det(C)$$
But the induction will be on the size of the matrix $I$.
Case $n=1$: let consider $I_1=1$, the $1\times1$ identity matrix, then we develop the determinant of $\begin{pmatrix} I_1 & 0 \\ 0 & C  \end{pmatrix}$ by the first row.
$$\det\begin{pmatrix} I_1 & 0 \\ 0 & C  \end{pmatrix}=(-1)^{1+1}(1)\det(C)=\det(C)$$
Case $n\to n+1$: suppose $$\det\begin{pmatrix} I_n & 0 \\ 0 & C  \end{pmatrix}=\det(C)$$
We will show it is true for $n+1$ by developping the determinant using the first row.
$$\det\begin{pmatrix} I_{n+1} & 0 \\ 0 & C  \end{pmatrix}=(-1)^{1+1}(1)\det\begin{pmatrix} I_n & 0 \\ 0 & C  \end{pmatrix}=\det(C)$$
Which end our induction proof.
A similar induction proof could be use to show that 
$$\det\begin{pmatrix} A & B \\ 0 & I  \end{pmatrix}=\det(A)$$
