$p$-part of cyclotomic character Let $K$ be a number field, $\bar K$ a separable closure and $p$ a rational prime and assume that $p \not= char(K)$. 
Consider the extension 
$$K(\mu_{p^\infty}) \mid K$$
which is obtained by adjoining roots of unity of $p$-power order to $K$. 

Why is the Galois group of this extension canonically isomorphic to the image of the $p$-part of the cyclotomic character,
  $$\chi_p: \text{Gal}(\bar K \mid K) \to \mathbb{Z}_p^\times \ \ ?$$

For clarity, the cyclotomic character is given by $g(\zeta) = \zeta^{\chi_p(g)}$ for $\zeta \in \mu_{p^\infty}$ and $g \in \text{Gal}(\bar K \mid K)$.
Thank you :-)
 A: Say $L/M/K$ is a tower for which $L$ and $M$ are Galois over $K$, then $L/M$ is automatically a Galois extension too. There is a canonical projection of Galois groups  $G(L/K)\to G(M/K)$, where $\sigma$ is sent to the restriction $\sigma|_M$. This makes sense because $\sigma M=M$ for all actions which pointwise fix $K$ on fields containing $M$ (this follows from $M$'s being normal over $K$).
Therefore, as $\overline{K}/K(\mu_{p^\infty})/K$ is such a tower, there is a projection $G(\overline{K}/K)\to G(K(\mu_{p^\infty})/K)$. Let's assume that $K$ is a global field with characteristic $\ne p$. Then a Galois action $\sigma$ on $K(\mu_{p^\infty})$ over $K$ is determined by the exponents $a_k$ in $\sigma\zeta_{p^k}=\zeta_{p^k}^{a_k}$. Conversely such a sequence $(a_1,a_2,\cdots)$ defines a Galois action if and only if it is coherent mod $p^k$ (i.e. $a_k\equiv a_{k+1}$ mod $p^k$) and $a_i\equiv1$ mod $p^m$ where $m$ is maximal for which $\mu_{p^m}\subseteq K$ (it is possible for $K$ to have $p$-power roots already, but only finitely many of them because it is a finite extension of its prime subfield). If $m=0$ there is no $\equiv1$ condition.
These sequences can be identified with elements of $1+p^m\Bbb Z_p$ under multiplication, or $\Bbb Z_p^\times$ if $m=0$. 
This approach can be abstracted into a categorical argument with limits. Suppose further $K$ contains no $p$th roots of unity. Then, since $G(\varinjlim L/K)=\varprojlim G(L/K)$, we have the following computation:
$$\begin{array}{ll} G(K(\mu_{p^\infty})/K) & =G(\varinjlim K(\mu_{p^n})/K) \\ & =\varprojlim G(K(\mu_{p^n})/K) \\ & =\varprojlim(\Bbb Z/p^n\Bbb Z)^\times \\ & =(\varprojlim\Bbb Z/p^n\Bbb Z)^\times \\ & =\Bbb Z_p^\times.\end{array}$$
You can do the same for when $K$ has $p$-power roots of unity, only there isn't good notation for the subgroup of $(\Bbb Z/p^n\Bbb Z)^\times$ of elements congruent to $1$ mod $p^m$ to work with that doesn't already give away the fact that $\Bbb Z_p$ is going to enter the fold. 
So the Galois groups are precisely the higher unit groups: $G(K(\mu_{p^\infty})/K)\cong U_m$.
