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.
 A: 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.
A: Using the definition of the determinant you have mentioned, you can show that every non-zero term in the summation for the left side lies in that of the right, and vice versa.
Look at the subpermutation $(\lambda_{m+1}, \ldots, \lambda_n)$ of the permutation on $\{1,\ldots, n\}$. Note that if any for any $m<i\leq n$, $\lambda_i \neq i$, then $a_{i,\lambda_i} = 0$, and hence the corresponding term will be zero.
Hence all the non-zero terms of the left summation have the permutation, when restricted to the last $n-m$ entries, equal to the identity. This also means that the sign of the permutation only depends on the action on the first $m$ terms.
Since it's obvious that every term in the summation of the right side corresponds to the term on the left side with the same permutation on $\{1,\ldots,m\}$ and identity on $\{m+1,\ldots,n\}$, you have that the two summations are equal. Hence the determinants are equal.
A: I strongly believe that the best way to show my indebtedness is to write out the proof. 
Without (Milind's) assistance I would not have completed it. 
$\begin{align} 
Given; 
D^{/}= 
\begin{vmatrix}
a_{11} &  \cdots& a_{1m} & 0 & \cdots & 0 \\
\cdot & \cdots & \cdot & \cdot & \cdots & \cdot \\
a_{m1} & \cdots & a_{mm} & 0 & \cdots & 0 \\
0 & \cdots & 0 & 1 & \cdots & 0 \\
\cdot & \cdots & \cdot & \cdot & \ddots & \cdot \\
0 & \cdots & 0 & 0 & \cdots & 1\\
\end{vmatrix}. 
\end{align}$
$\begin{align} 
To prove; 
D^{/}=
\begin{vmatrix}
a_{11} &  \cdots& a_{1m} \\
\cdot & \cdots & \cdot \\
a_{m1} & \cdots & a_{mm} \\
\end{vmatrix}. 
\end{align}$
Proof;
$\begin{align}
{a_{ij}^/} = 
\begin{cases}
a_{ij};\qquad i=j:1\leq\ j\le\ m \\[2ex]
a_{ij} = 0; i\neq\ j: m\lt j \le n\\[2ex]
a_{ij} = 1; i=j: m+1\le\ j\le\ n; WHY?- (Hypothesis).
\end{cases}
\end{align}$
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 $\begin{align}
\begin{vmatrix}
a_{11} & a_{21} & \cdots & a_{1m}\\
a_{21} & a_{22} & \cdots & a_{2m}\\
\cdot & \cdot & \cdots & \cdot\\
a_{n1} & a_{n2} & \cdots  & a_{nn}\\
\end{vmatrix}, \text{or more briefly, by}
\end{align}$
$\begin{align}
{\begin{vmatrix}
a_{ij}\\
\end{vmatrix}_n}; (Reason [1]).
\end{align}$
$\begin{align} D^{/}= \sum_{(\lambda_1, \cdots , \lambda_n)} {\epsilon (\lambda_1, \cdots , \lambda_n)} a_{1\lambda_1}^/ ,\cdots, a_{n\lambda_n}^/; WHY?- (Reason [1]).\end{align}$
$\begin{align} D^{/}= \sum_{(\lambda_1, \cdots , \lambda_n)} {\epsilon (\lambda_1, \cdots ,\lambda_m, \lambda_{m+1}, \cdots, \lambda_n)} a_{1\lambda_1}^/ ,\cdots, a_{m\lambda_m}^/, a_{{m+1}\lambda_{m+1}}^/, \cdots, a_{n\lambda_n}^/ \end{align}$
For any $j$, where $i=j:1\leq\ j\le\ m$, we observe that $a_{ij} = a_{ij}^/$.
It follows immediately that
$\begin{align} D^{/}= \sum_{(\lambda_1, \cdots , \lambda_n)} {\epsilon (\lambda_1, \cdots , \lambda_n)} a_{1\lambda_1} ,\cdots, a_{n\lambda_n}; WHY?- (Hypothesis).\end{align}$
$\begin{align} D^{/}= \sum_{(\lambda_1, \cdots , \lambda_n)} {\epsilon (\lambda_1, \cdots ,\lambda_m, \lambda_{m+1}, \cdots, \lambda_n)} a_{1\lambda_1} ,\cdots, a_{m\lambda_m}, a_{{m+1}\lambda_{m+1}}, \cdots, a_{n\lambda_n} \end{align}$
By looking at the sub-permutation $(\lambda_{m+1}, \cdots, \lambda_n)$ of the longer permutation $(1, \cdots, n)$ we immediately observe that for any $j$, where $i\neq\ j: m\lt j \le n$, $a_{ij} = 0$. Hence the corresponding term will be zero ; WHY?- (Hypothesis).
And in consequence, for any $j$, when $i=j$ and $m+1\le\ j\le\ n$, $a_{ij} = 1$; the non-zero terms which are left in the summation have their permutation, when restricted to the last $n-m$ entries, equal to the identity. This clearly implies that the sign of the permutation is dependent only on the action of the first $m$ terms; WHY?- (Hypothesis).
Hence it follows from the previous results we obtained that
$\begin{align}D^{/}= \sum_{(\lambda_1, \cdots , \lambda_m)} {\epsilon (\lambda_1, \cdots , \lambda_m)} a_{1\lambda_1} ,\cdots, a_{m\lambda_m}\end{align}$
$\begin{align} 
D^{/}=
\begin{vmatrix}
a_{11} &  \cdots& a_{1m} \\
\cdot & \cdots & \cdot \\
a_{m1} & \cdots & a_{mm} \\
\end{vmatrix}
\end{align}$;WHY?- (Reason [1]).
Q.E.D.
This completes the proof of Exercise 1.3.1.
