Why is the group of principal units of a local field uniquely divisible by $n$? I am reading a proof with the followings setup and claim.
$K/F$ is a Galois extension of local fields with group $G$ of order $n = q^s$, where $q$ is prime and $s \geq 1$. Assume the maximal ideal $\mathfrak{p}_F$ in $\mathcal{O}_F$ totally ramifies in $K$ (so that $\mathfrak{p}_F\mathcal{O}_K = \mathfrak{p}^n$ where $\mathfrak{p}$ is the maximal ideal in $\mathcal{O}_K$), and $\mathfrak{p}_F \nmid n$.
If $U = \mathcal{O}_K^\times$ and $U^{(n)} = 1 + \mathfrak{p}^n$, then the author states that "$U^{(1)}$ is the inverse limit of finite $p$-groups, $\mathfrak{p}_F | p$, so is uniquely divisible by $n$", which is the statement that I can't understand. I see that $$U^{(1)} = \varprojlim_{n > 1} U^{(1)}/U^{(n)}.$$ Since $U/U^{(n)} \cong (\mathcal{O}_K/\mathfrak{p}^n)^\times$, I guess it would suffice to show that these are $p$-groups for $n > 1$, because $U^{(1)}/U^{(n)} \leq U/U^{(n)}$. But I'm not entirely sure why that's true.
But even then, I'm not sure why it follows that $U_1$ is uniquely divisible by $n$, which is prime to $p$. Isn't it true that in general, a finite $p$-group isn't uniquely divisible, let alone divisible, by $n$?
In any case, the point of this is to show that $H^1(G, U^{(1)}) = 0$, so if there's another elementary way of seeing this, that would help as well.
What am I missing here?
 A: Finite $p$-groups are uniquely divisible by any $n$ relatively prime to $p$. The point is that in a finite $p$-group $G$ the map
$$\mathbb{Z} \ni n \mapsto (g \mapsto g^n)$$
extends uniquely to a map from the $p$-adic integers $\widehat{\mathbb{Z}}_p$, and in the $p$-adic integers any $n$ relatively prime to $p$ has a unique multiplicative inverse. (I could use a finite quotient of $\mathbb{Z}$ depending on the size of $G$, but this way of saying things is independent of the size of $G$.) 
For example, in a finite $2$-group there are unique cube roots given by raising an element to the power of
$$\frac{1}{3} = 1 - 2 + 2^2 - 2^3 \pm \dots$$
A: Here’s another approach, unnecessarily recondite. Let $z$ be a $p$-adic integer. We know that $z$ is the $p$-adic limit of positive ordinary integers $n_j$. Now look at the binomial expansion $(1+t)^{n_j}=1+\sum_{i=1}^\infty c_{ij}t^i$. Each coefficient, say of $t^i$, is $C_i(n_j)$, where
$$
C_i(x)=\frac{x(x-1)(x-2)\cdots(x-i+1)}{i!}\,.
$$
From its polynomial description, you know that $C_i$ is a $p$-adically continuous function, and it takes ordinary integers to ordinary (and so $p$-adic) integers, so that the value
$$
\lim_{j\to\infty}C_i(n_j)\,,
$$
being a continuous function applied to a convergent $p$-adic sequence, has a valid $p$-adic limit. In other words,
$$
(1+t)^z\in\Bbb Z_p[[t]]\,.
$$
This means that (when you plug in an element $\alpha$ of the maximal ideal of any $p$-adic ring for $t$), $(1+\alpha)^z$ makes sense. In particular, if $m$ is prime to $p$, $1/m$ is a $p$-adic integer, so you have $(1+\alpha)^{1/m}$ as a principal unit.
