We defined the wedge-product as follows: $ w \wedge \mu = \frac{(k + l)!}{k!l!} Alt (w \otimes \mu) $ (here you'll find additional information).
In the context of a proof we came to a point where we used a permutation to change the order of the inserted values of the wedge product. This permutation was defined as $ \tau:= \begin{cases} x- m, & \mbox{for } x>m \\ x+k, & \mbox {in all other cases} \end{cases}$
Then we had the sum (where f is a k-form and g is a m-form)
$\frac{1}{k! m!} \sum_{\sigma \in \Sigma _ {k + m}} sign(\sigma) g _\sigma (v _ {k +1},...,v _ {k +m}) f _\sigma (v _ {1},...,v _ {k })$ (1)
where the $\sigma$ in the index shows that the argument values for f and g are $v_{\sigma(i)}$.
If we now use $\tau$ the following sum shall result:
$\frac{1}{k! m!} \sum_{\sigma \in \Sigma _ {k + m}} sign(\sigma ) g _{\sigma \circ \tau} (v _ {1},...,v _ {m}) f _{\sigma \circ \tau} (v _ {m+1},...,v _ {m+k })$ (2)
May someone explain me, why (1) equals (2) and the value isn't changed by the use of $\tau$?
I attached in the edit the whole claculation:
