I need help with the following:
Let $K$ be a field. Let $\sigma \in S_n$ be a permutation of numbers $1, \dots, n$ and let $f:K^n \rightarrow K^n, \begin{pmatrix} x^1 \\ \vdots \\ x^n \end{pmatrix} \mapsto \begin{pmatrix} x^{\sigma(1)} \\ \vdots \\ x^{\sigma(n)} \end{pmatrix}$ be a linear mapping.
(i) Calculate the linear transformation matrix $A_\sigma$ of $f$ regarding the standard basis $\{e_1,\dots,e_n\}$ of $K^n$.
(ii) Show that $\mathrm{det}(A_\sigma) = \mathrm{sgn}(\sigma)$.
(iii) Show that $A^{n!} = 1\!\!1$, $1\!\!1 := \begin{pmatrix} 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{pmatrix}$.
Ideas:
(i) $A_\sigma = \begin{pmatrix} e_{\sigma(1)}^T \\ \vdots \\ e_{\sigma(n)}^T \end{pmatrix} = \begin{pmatrix} e_{\sigma(1)} & \cdots & e_{\sigma(n)}\end{pmatrix}$. I'm not sure if I need to show the first equality and how to do it for the second?
(ii) Let $a^i_j$ where $1 \leq i,j \leq n$ be the $i$th row, $j$th column of matrix $A_\sigma$. We know that $\mathrm{det}(A_\sigma) = \sum_{\pi \in S_n} \mathrm{sgn}(\pi) \cdot a_1^{\pi(1)} \cdot a_2^{\pi(2)} \cdot \ldots \cdot a_n^{\pi(n)}$. [Leibniz formula]
In (i) we see that the $j$th column of $A_\sigma$ is the unit vector $e_{\pi(j)}$, so that $a_1^{\pi(1)} \cdot a_2^{\pi(2)} \cdot \ldots \cdot a_n^{\pi(n)} = 1$ and thus $\mathrm{det}(A_\sigma) = \sum_{\pi \in S_n} \mathrm{sgn}(\pi)$. I don't know how to go from here...
(iii) Observation: Let $d$ be the number of columns not matching the identity matrix $1\!\!1$, then $A^d = 1\!\!1$.