I'm currently learning Exterior Algebra to get a proof of Stoke's Theorem. Unfortunately, I don't understand the proof of a certain statement. Before that I show you the proof, I have to introduce you some to some definition.
By knowing that $\alpha \in L^r(E,E)$, and $\epsilon(\sigma)$ is the signature of the corresponding permutation. $$Permutation \space Operator:P_\sigma (\alpha)(v_1,\ldots, v_r):= \alpha (v_{\sigma(1)},\ldots, v_{\sigma(r)})$$ $$Antisymmetric \space Operator:A(\alpha)(v_1,\ldots, v_r):=\frac{1}{r!}\sum_{\sigma\in S_r}\epsilon(\sigma)P_\sigma(\alpha)(v_1,\ldots, v_r)$$
We want to prove this statement. $$(A \circ P_\sigma) (\alpha) (v_1,\ldots, v_r)=(P_\sigma \circ A)(\alpha) (v_1,\ldots, v_r)=\epsilon(\sigma)A(\alpha) (v_1,\ldots, v_r)$$ Here is the demonstration written in a compact form. $$(P_\sigma \circ r!A) (\alpha)=\color{red}{\sum_{\tau \in S_r} \epsilon(\tau)P_\sigma \circ P_\tau (\alpha)=\sum_{\tau \in S_r} \epsilon(\sigma^{-1}\tau)P_\tau (\alpha)}=\sum_{\tau \in S_r} \epsilon(\sigma^{-1})\epsilon(\tau)P_\tau(\alpha)=\epsilon(\sigma)\sum_{\tau \in S_r}\epsilon(\tau)P_\tau(\alpha)=\epsilon(\sigma)r!A (\alpha)$$
The only thing that I don't understand is where the equality colored in red comes from. I would be really glad to get an easy answer, such that someone like me who doesn't have some solid knowledge in group theory can understand it.
Thanks in advance for your support.