How to prove the following exercise by using the definition of a determinant? $\begin{align} \begin{vmatrix}
a_{11} &  \cdots& a_{1m} & 0 & \cdots & 0 \\
\cdot & \cdots & \cdot & \cdot & \cdots & \cdot \\
a_{m1} & \cdots & a_{mm} & 0 & \cdots & 0 \\
0 & \cdots & 0 & 1 & \cdots & 0 \\
\cdot & \cdots & \cdot & \cdot & \ddots & \cdot \\
0 & \cdots & 0 & 0 & \cdots & 1
\end{vmatrix} =
\begin{vmatrix}
a_{11} &  \cdots& a_{1m} \\
\cdot & \cdots & \cdot \\
a_{m1} & \cdots & a_{mm} \\
\end{vmatrix}. 
\end{align}$
i.e. Definition of a determinant; 
The determinant of the array A is the number 
$$ \sum_{(\lambda_1, \cdots , \lambda_n)} {\epsilon (\lambda_1, \cdots , \lambda_n)} a_{1\lambda_1} ,\cdots, a_{n\lambda_n} $$ where the summation extends over all n! arrangements $(\lambda_1,\cdots ,\lambda_n)$ of $(1, \cdots, n)$. This determinant is denoted by 
$\begin{align}
\begin{vmatrix}
a_{11} & a_{21} & \cdots & a_{1m}\\
a_{21} & a_{22} & \cdots & a_{2m}\\
\cdot & \cdot & \cdots & \cdot\\
a_{n1} & a_{n2} & \cdots  & a_{nn}\\
\end{vmatrix}, \text{or more briefly, by}
\end{align}$
$\begin{align}
{\begin{vmatrix}
a_{ij}\\
\end{vmatrix}_n}.
\end{align}$
$\begin{align}
A = 
\begin{matrix}
a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\cdot & \cdot &\cdots & \cdot\\
a_{n1} & a_{n2} & \cdots & a_{nn}\\
\end{matrix}
\end{align}$
This exercise is taken from the book 'An Introduction to Linear Algebra' by L. Mirsky.
Page number 13, exercise 1.3.1.

Use the fact that the determinant of a block matrix is equal to the product of the determinants of matrix $A$ and $B$. 
So $$\begin{vmatrix} A & 0 \\ 0 & B\end{vmatrix} = \det(A) \det(B)$$ 
Here, $A$ is an $m\times m$ matrix, $B$ is a $(n - m)\times (n - m)$ matrix.
This can be proven by induction.
