In the Wikipedia article on alternating multilinear forms, two equivalent definitions of the exterior products are given.
Definition 1: $$ \omega \wedge \eta ={\frac {(k+m)!}{k!\,m!}}\operatorname {Alt} (\omega \otimes \eta ),\tag{1} $$ where $$ \operatorname{Alt}(\omega)(x_1,\ldots,x_k)=\frac{1}{k!}\sum_{\sigma\in S_k}\operatorname{sgn}(\sigma)\,\omega(x_{\sigma(1)},\ldots,x_{\sigma(k)});\tag{1.5} $$ Definition 2: $$ {\omega \wedge \eta(x_1,\ldots,x_{k+m})} = \sum_{\sigma \in Sh_{k,m}} \operatorname{sgn}(\sigma)\,\omega(x_{\sigma(1)}, \ldots, x_{\sigma(k)}) \eta(x_{\sigma(k+1)}, \ldots, x_{\sigma(k+m)}),\tag{2} $$ where here $Sh_{k,m} ⊂ S_{k+m}$ is the subset of $(k,m)$ shuffles: permutations $σ$ of the set $\{1, 2, ..., k + m\}$ such that $σ(1) < σ(2) < ... < σ(k)$, and $σ(k + 1) < σ(k + 2) < ... < σ(k + m)$.
Note that (1) can be rewritten as $$ {\omega \wedge \eta(x_1,\ldots,x_{k+m})} = \frac{1}{k!m!}\sum_{\sigma \in S_{k+m}} \operatorname{sgn}(\sigma)\,\omega(x_{\sigma(1)}, \ldots, x_{\sigma(k)}) \eta(x_{\sigma(k+1)}, \ldots, x_{\sigma(k+m)})\tag{1'} $$ Thus to show that (1) and (2) are equivalent, it suffices to show that (1') are (2) are the same. Ignoring the $\operatorname{sgn}(\sigma)$ part, one can count the number ofterms in (1') and (2).
Here is my question:
How can one show that (1') and (2) are the same?