I've tried show $\det \left( \begin{matrix} \delta_{i_{\sigma(1)}}^{i_{1}} & \cdots & \delta_{i_{\sigma(k)}}^{i_{1}} \\ \vdots & \ddots & \vdots \\ \delta_{i_{\sigma(1)}}^{i_{k}} & \cdots & \delta_{i_{\sigma(k)}}^{i_{k}} \end{matrix}\right) = \operatorname{sgn}(\sigma)$,
Notation: $\delta_{m}^{n} = 1$ when $m = n$, $0$ when $m \neq n$.
Where $\sigma \in S_k = \{\alpha:\{1,\cdots,k\} \rightarrow \{1,\cdots,k\} : \alpha$ bijective $\}$, $i_1,\cdots, i_k \in \mathbb{N}$ s.t. $i_s \neq i_t, \forall s,t \in \{1,\cdots,k\}$, and $\operatorname{sgn}(\sigma) = \pm 1$, $1$ when $\sigma$ is even and $-1$ when $\sigma$ is odd.
My attempt: determinant is multi-linear and alternating, so in each $s$-line we have all entries $0$ except the entry correspondent to column $\sigma^{-1}(s)$, in this entry $\delta^{i_s}_{i_{\sigma(\sigma^{-1}(s))}} = \delta_{i_s}^{i_s} = 1$. I've tried use the alternating of determinant and this fact to show that when $\sigma$ is even we have even changes of columns to transform the matrix $\left( \begin{matrix} \delta_{i_{\sigma(1)}}^{i_{1}} & \cdots & \delta_{i_{\sigma(k)}}^{i_{1}} \\ \vdots & \ddots & \vdots \\ \delta_{i_{\sigma(1)}}^{i_{k}} & \cdots & \delta_{i_{\sigma(k)}}^{i_{k}} \end{matrix}\right)$ to identity in $M_k(\mathbb{R})$, analogous when $\sigma$ is odd. But I can't show this.
Thanks!
Rewrite your matrix as a permutation of the columns of the identity matrix, $\det\left(C_{\sigma(1)},\ldots,C_{\sigma(k))}\right)$ then the result follows from multi-linear and alternating properties.