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?