Operator action $(A_{1}\wedge \ldots \wedge A_n)(e_1\wedge \ldots \wedge e_n)=\sum\limits_{\sigma \in S_{n}}(-1)^\sigma A_{1}(e_{\sigma(1)})\wedge \ldots \wedge A_{n}(e_{\sigma(n)})$
Is this definition or can you somehow prove it?
 A: The wedge product does not act on another wedge product. If the $A_i$ are covectors on a vector space $V$, then they act on elements of $V$, and so the wedge of $n$ of them acts on $n$ elements of $V$. Otherwise, your expression is almost correct, but with a similar mistake on the other side of the equation. The wedge of $n$ co-vectors is an alternating $n$-linear form, so it doesn't make sense to have a wedge in the final answer; it should be a number. The correct expression is \begin{align*}
&(A_1\wedge A_2\wedge\cdots\wedge A_n)(e_1,e_2,\cdots, e_n)\\
&=\sum\limits_{\sigma\in S_n}(\text{sgn}(\sigma))A_1\left(e_{\sigma(1)}\right)A_2\left(e_{\sigma(2)}\right)\cdots A_n\left(e_{\sigma(n)}\right).\end{align*} 
This is the definition (or, at least, follows from the definition of the wedge of two covectors). 
Also, note that it is, in fact the determinant of $[A_i(e_j)]$. I should also note that we usually write the subscript that you used on the $A$'s as superscripts i.e. $$(A^1\wedge A^2\wedge\cdots\wedge A^n)(e_1,e_2,\cdots,e_n)=\det [A^i(e_j)].$$
