The image of cyclotomic character is open? Let $K$ be a finite extension of $\mathbb{Q}_p$, and define $t=(1,\varepsilon_1,\varepsilon_2,\varepsilon_3...)\in \bar{K}_\mathbb{N}$ where $\varepsilon_i^{p^i}=1$ and $\varepsilon_{i+1}^p=\varepsilon_i$ for any $i\in\mathbb{N}$, then we can define a group homomorphism $\text{Gal}_{\bar{K}/K}\xrightarrow{\chi} \mathbb{Z}_p^*$ which is defined by $g\circ t=t^{\chi(t)}$. If $x=\sum_{i\geq0} a_ip^i\in\mathbb{Z}_p^*$, then we define $t^x=(1,\varepsilon_1^{a_1},\varepsilon_2^{a_1+a_2p},...)$.
Then the question is how to see $\text{Im}(\chi)$ is an open subgroup of $\mathbb{Z}_p^*$. I think we can prove that there exists a positive integral number $n$ such that $1+p^n\mathbb{Z}_p\subseteq \text{Im}(\chi)$. But how to choose this $n$?
Thanks!
 A: Hint: Since $K$ is a finite extension of $\mathbb Q_p$, there is some $n$ such that $K$ contains the $p^{n-1}$-th roots of unity but not the $p^n$-th ones.
Now how does $Gal(\bar K\vert K)$ act on the $p^k$-th roots of unity for $k \ge n$?
A: Torsten answer is good, but I wanted to say your notations are not adapted. With $\zeta_{p^r}$ the largest $p$-root of unity in $K(\zeta_p)$ then $$Gal(K(\zeta_{p^\infty})/K(\zeta_p))=Gal(\Bbb{Q}_p(\zeta_{p^\infty})/\Bbb{Q}_p(\zeta_{p^r}))= 1+p^r \Bbb{Z}_p\subset \Bbb{Z}_p^\times$$
where $a\in  1+p^r \Bbb{Z}_p$ gives the automorphism $\zeta_{p^n}\to \zeta_{p^n}^{a\bmod p^n}$. 
Of course $Gal(K(\zeta_{p^\infty})/K(\zeta_p)$ is of finite index in $Gal(K(\zeta_{p^\infty})/K)\subset Gal(\Bbb{Q}_p(\zeta_{p^\infty})/\Bbb{Q}_p)=\Bbb{Z}_p^\times$.
The cyclotomic character of $Gal(\overline{K}/K)$ is just sending $\sigma$ to $\sigma|_{K(\zeta_{p^\infty})}$ to $\Bbb{Z}_p^\times$.
Also that $\zeta_{p^n}^{p^n}=1$ is not enough, you need $\zeta_{p^n}^p = \zeta_{p^{n-1}}$.
