How can commuting with Frobenius imply the order of an element in the inertia group.

Question

In this video, one asserts, in the beginning, that, for $\tau\in \mathbb V_0$ such that $\tau$ generates $V_0/V_1$ in the quotient group, and $\sigma\in \mathbb Z$ which is a Frobenius in $\mathbb Z/\mathbb T$, if $\tau\sigma=\sigma\tau$, then $\tau^{q-1}=\iota$ where $q$ is the order of the finite residue field $\kappa$. I have no idea how to show this. Any hints are mostly appreciated, thanks in advance.
P.S. $\mathbb V_i$ are ramifications groups of a number field $\mathbb k$ with some prime, and $\mathbb Z$ the decomposition group of that prime in $\mathbb k$, with $\mathbb T=\mathbb V_0$. And notice that $q$ is the order of the residue field $\kappa=\mathbb k/\mathfrak p$, not of the finite field extension $\mathbb K^0=\mathbb K/\mathfrak P$.
Edit:
This video is some kind of blurred and, for the dearth of time, cannot give a complete proof of the theorem. Though I can understand the proof by CFT, I found it quite difficult to understand those lemmas. And any reference to the paper the lecturer uses is rather appreciated. Thanks.
Another Edit:
I think that, if we view $\tau$ as a representation, then the commuting condition becomes sort of saying that the representation is one-dimensional, so that it is contained in the ground field, fromp which follows the order-statement. Are there any errors? Thanks again.