Determinant of $\delta$ function Let 
$$\delta_i^j=\left\{
\begin{aligned}
1 ~~~~~~i=j   \\
0 ~~~~~~i\ne j
\end{aligned}
\right.
$$
$1\le i,j\le n$.
How to prove 
$$
\begin{vmatrix}
\delta_{j_1}^{i_1} ~...~ \delta_{j_n}^{i_1} \\
\\
\delta_{j_1}^{i_n} ~...~ \delta_{j_n}^{i_n}
\end{vmatrix}
=\left\{
\begin{aligned}
1          &~~~~~~~~~\forall~ 1\le k,l \le n , i_k\ne i_l \text{ and $j_1...j_n$ is even permutation of $i_1...i_n$ }    \\
-1         &~~~~~~~~~\forall~ 1\le k,l \le n , i_k\ne i_l \text{ and $j_1...j_n$ is odd permutation of $i_1...i_n$ }     \\
0          &~~~~~~~~~\text{others}     \\
\end{aligned}
\right.
$$
I try to use induction to prove it ,but seemly it is too complex. 
 A: Denote the given matrix by $D$. If there are $k \ne l$ such that $i_k = i_l$, then the $k$-th and $l$-th row of $D$ are equal, hence $\det(D) = 0$. So suppose all $i_k$ are distinct. If $i_k = j_k$ for all $k$, then $D$ is the identity matrix, so $\det(D) = 1$. For the general case, we first prove the following: if we switch $j_k$ and $j_l$ for some $l \ne k$, then the determinant changes sign. Indeed, switching $j_l$ and $j_k$ corresponds to switching the $j$-th and $k$-th column of $D$. Hence, the determinant switches sign.
The statement follows now from the following observation: denote by $\sigma$ the permutation of $i_1, \ldots, i_n$ resulting in $j_1, \ldots, j_n$, i.e. $\sigma(i_k) = j_k$. Writing $\sigma$ as a composition of transpositions, we obtain $D$ from the identity matrix by consecutevily switching the columns corresponding to the transposition. With each transposition, the determinant flips sign. In the end, the determinant equals the sign of $\sigma$, which is $1$ if $\sigma$ is an even permutation and $-1$ if $\sigma$ is odd.
