I am studying Serre's book Local Fields on ramification groups for cyclotomic extensions of $\mathbb{Q}_p$ (Section 4.4). Let $K_n = \mathbb{Q}(\zeta)$ be a totally ramified extension over $\mathbb{Q}_p$, where $\zeta$ is a $p^n$-th root of unity. Denote by $\pi = \zeta-1$ a uniformizer of $K_n$ and by $v_\pi$ the valuation.
According to Serre's result (Prop18), $i_G(s)$ only takes the values in $\{p^i\}_{0\le i \le n-1}$, where $i_G(s):= v_{\pi}(s(\zeta)-\zeta)$ for $s\in Gal(K_n/\mathbb{Q}_p)= (\mathbb{Z}/p^n\mathbb{Z})^\times $. I am not sure if I have understood it correctly.
Now take $a = 1 + 2p\in (\mathbb{Z}/p^n\mathbb{Z})^\times$, which corresponds to $s:\zeta\mapsto \zeta^a$ in the Galois group. Since $\pi = \zeta - 1$, we would have $$ s(\zeta) - \zeta = \zeta((1+\pi)^{2p} - 1) = \zeta(\pi^{2p} + \text{parts divisible by } p). $$ I suppose this would suggest $v_\pi(s(\zeta)-\zeta) = 2p$ if $p$ and $n$ are large? But it is not in $\{p^i\}_{0\le i \le n-1}$. What did I do wrong?
With $s: \zeta \mapsto \zeta^{2p+1}$ where $\zeta=\zeta_{p^{r+2}}$, when computing $$v(s(\pi)-\pi) = v(\zeta ((1+\pi)^{2p}-1))$$ You forgot the $$\pi^p {2p\choose p}$$ term whose valuation is $p\, v(\pi)$