Show $x_{σ(1)}∧x_{σ(2)}∧\cdots∧x_{σ(k)}=\operatorname{sgn}(σ)x_1∧x_2∧\cdots∧x_k$ Im currently learning exterior algebra, and I found in wikipedia (https://en.wikipedia.org/wiki/Exterior_algebra#Alternating_product) the following problem:

If $σ$ is a permutation of the integers $\{1, \ldots, k\}$, and $x_1, x_2,\ldots, x_k$ are elements of $V$, it follows that
$$ x_{\sigma (1)}\wedge x_{\sigma (2)}\wedge \cdots \wedge x_{\sigma (k)}=\operatorname{sgn} (\sigma )x_1\wedge x_2\wedge \cdots \wedge x_k$$

I know what a permutation is, but I don't know how to prove that problem. Any idea or help will be appreciated.
 A: By definition
$$
x_{1}\wedge x_{2}\wedge \cdots \wedge x_{k}=\sum_{\pi\in S_n}\operatorname {sgn} (\pi )x_{\pi(1)} \otimes \cdots \otimes x_{\pi(k)}
$$
where $S_n$ is the group of permutation.
(Sometimes there is a pre-factor $1/n!$ in the definition). Then
$$
 x_{\sigma (1)}\wedge x_{\sigma (2)}\wedge \cdots \wedge x_{\sigma (k)}=\sum_{\pi\in S_n}\operatorname {sgn} (\pi )x_{\pi(\sigma(1))}\otimes \cdots \otimes x_{\pi(\sigma(k))}=\\
\operatorname {sgn} (\sigma)\cdot \sum_{\pi\in S_n}\operatorname {sgn} (\sigma)\cdot \operatorname {sgn} (\pi )x_{\pi(\sigma(1))}\otimes \cdots \otimes x_{\pi(\sigma(k))}=\\
\operatorname {sgn} (\sigma)\cdot \sum_{\pi\in S_n}\operatorname {sgn} (\pi\circ \sigma )x_{\pi\circ \sigma(1)}\otimes \cdots \otimes x_{\pi\circ \sigma(k)}
$$
Now, for any $\pi$, define $\pi'=\pi\circ \sigma$. Then
$$
 x_{\sigma (1)}\wedge x_{\sigma (2)}\wedge \cdots \wedge x_{\sigma (k)}=\operatorname {sgn} (\sigma)\cdot \sum_{\pi\in S_n}\operatorname {sgn} (\pi' )x_{\pi'(1)}\otimes \cdots \otimes x_{\pi'(k)}=\\
=\operatorname {sgn} (\sigma)\cdot \sum_{\pi'\in S_n}\operatorname {sgn} (\pi' )x_{\pi'(1)}\otimes \cdots \otimes x_{\pi'(k)}=\\
\operatorname {sgn} (\sigma) \cdot x_{1}\wedge x_{2}\wedge \cdots \wedge x_{k}
$$
