TITLE: The torsion subgroup of principal units $U^{(1)}$
QUESTION [5 upvotes]: $\newcommand{\U}{U^{(1)}}$
$\newcommand{\O}{\mathcal{O}}$
$\newcommand{\p}{\mathfrak{p}}$
$\DeclareMathOperator{\char}{char}$
$\newcommand{\N}{\mathbb{N}}$
I have a question about the torsion subgroup of principal units $\U$.
Let $K$ be a local field with valuation ring $\O$, maximal ideal $\p$.
Let $q = p^f = \#\O/\p$. Let $\char K = 0$. The group of principal units $\U$ is
$$
\U := 1 + \p.
$$
For $n \in \N$, let $\mu_n$ be a group of $n$-th roots of unity,
i.e.
$$
  \mu_n := \{ x \in K \mid x^n = 1 \}.
$$
Then, I want to show that:

The torsion subgroup of $\U$ can be written as $\mu_{p^a}$ for some $a \in \mathbb{N}$.

First, I tried to show that for any $x \in \U$ which has finite order,
an order of $x$ can be written as $p^n$ for some $n \in \N$,
but I failed.
This question is related to the Proposition (5.7) in
Neukirch, "Algebraic Number Theory" at page 140.

REPLY [4 votes]: Suppose $1+x$ is a torsion element of $U^{(1)}$, $x \in \mathfrak p$, killed by $z \in \mathbb Z_p$. That is, $(1+x)^z=1$. If $z=p^au$, $u \in \mathbb Z_p^{\times}$. Let $uv=1$, $v \in \mathbb Z_p^\times$.
$$(1+x)^{{p^a}u}=1 \implies (1+x)^{{p^a}uv}=(1+x)^{p^a}=1 \implies 1+x \in \mu_{p^a}.$$