I can do the first part, where $A$ is a $1\times1$ matrix.
$$\det(C) = \Sigma_i (-1)^{i+j} C_{ij} \det(F_{ij})$$
$F_{ij}$ is the co-factor matrix (a matrix that includes all the elements of $C$ except for the ith row and jth column)
If we set $j=1$ then the only value for $i$ for which $C_{ij}\neq0$ is 1. we also know that $C_{11}=Det(A)$ and therfoer the co-factor matrix is $F_{11}B$.
Our expression above becomes $Det(C)=Det(A)det(B)$ as required.
The second bit, however, I couldn't do via induction. I did find one method however:
$$ C=\begin{bmatrix} A & O \\ O & I_m \end{bmatrix} \begin{bmatrix} I_n & O \\ O & B \end{bmatrix} $$ where $I_n$ is the $n\times n$ identity matrix.
$$\det(C) = \det\begin{bmatrix} A & O \\ O & I_m \end{bmatrix} \det\begin{bmatrix} I_n & O \\ O & B. \end{bmatrix}$$
Which is clearly (using similar logic to above) $\det(C) = \det(A) \det(B)$
Can anyone please
tell me if my proof is valid and
show me how to solve this via induction?