Ring of integers and unit group for local fields On page 6 of Sutherland's notes (https://math.mit.edu/classes/18.785/2018fa/LectureNotes27.pdf), he claims that 
$$\mathcal{O}_K^\times\cong\mathbb{F}_p^\times\times\mathcal{O}_K.$$I understand the structure theorem (from Neukirch, for instance) tells us that $$\mathcal{O}_K^\times\cong\mathbb{F}_p^\times\times U^{(1)}.$$ In both cases, these are isomorphisms of topological groups. I don't quite see why $\mathcal{O}_K\cong U^{(1)}$ -- does it follow from some application of the p-adic logarithm (note that the first is an additive group and the latter is a multiplicative group).
 A: For $K$ a finite extension of $\Bbb{Q}_p$ of degree $N$ with residue field $O_K/(\pi)=\Bbb{F}_{p^f}$,
By Hensel lemma the roots of $x^{p^f-1}-1$ lift to $O_K$ and $$O_K^\times = \langle \zeta_{p^f-1}\rangle \times (1+\pi O_K)$$
Because $(1+\pi a)^{p^m}= 1+O(\pi^m)$ then $(1+\pi a)^b=\lim_{m \to \infty}(1+\pi a)^{b\bmod p^m}$ is well-defined for $b\in \Bbb{Z}_p$ so that

$1+\pi O_K$ is a $\Bbb{Z}_p$ module.

It is finitely generated because $1+p^3 O_K$ is of finite index and $\exp,\log$ make it isomorphic to $p^3 O_K\cong \Bbb{Z}_p^N$ 
Thus for some $r$, $(1+\pi O_K)_{tors}= \langle \zeta_{p^r}\rangle$ and 

$M=(1+\pi O_K)/\langle \zeta_{p^r}\rangle$ is a finitely generated torsion free $\Bbb{Z}_p$ module

whence $M\Bbb{Q}_p=M\otimes_{\Bbb{Z}_p}\Bbb{Z}_p[p^{-1}]$ is a $\Bbb{Q}_p$-vector space and taking $m\in M$ such that $m \Bbb{Q}_p\cap M =m\Bbb{Z}_p$ and repeating with $M'=M/m\Bbb{Z}_p$ and so on (as $M'\Bbb{Q}_p=M\Bbb{Q}_p/m\Bbb{Q}_p$ is $\dim(M\Bbb{Q}_p)-1$ dimensional)
we get that $M$ is a rank $N$ free $\Bbb{Z}_p$ module, there are some $\alpha_j\in O_K$ such that
$$M=\prod_{j=1}^N (1+\pi \alpha_j)^{\Bbb{Z}_p}\langle \zeta_{p^r}\rangle$$
and hence 
$$O_K^\times =\langle \zeta_{p^f-1}\rangle\times \langle \zeta_{p^r}\rangle \times \prod_{j=1}^N (1+\pi \alpha_j)^{\Bbb{Z}_p}\cong \langle \zeta_{p^r(p^f-1)}\rangle \times O_K$$

I don't know how to adapt this to the other kind of local fields : the finite extensions of $\Bbb{F}_p((x))$.

A: Even though you certainly didn’t ask about the case of characteristic $p$, the excellent answer of @reuns begs for the extension below.
The case of a finite extension of $\Bbb F_p((x))$ is nothing like the situation in characteristic zero. First, the only locally compact fields in positive characteristic are the $\Bbb F_q((x))$, $q$ being a power of $p$. That’s the easy part. Also, the units are still $\Bbb F_q^\times\times U^{(1)}$; I’ll write the second factor as $1+\mathfrak m$. But this latter is torsion-free, and thus a torsion-free $\Bbb Z_p$-module. But not finitely generated, as you’ll see.
A moment’s reflection tells you that the series $\{1+x^m\}$ for $p\nmid m$ all are $\Bbb Z_p$-independent, multiplicatively of course. And in the simplest case $q=p$, one gets
$$1+\mathfrak m\cong\prod_{p\nmid m}\Bbb Z_p\,,$$ that is, a product (not direct sum) of those infinitely many copies of $\Bbb Z_p$.
For the general case $q>p$, the story is more complicated, best illuminated by means of the dreaded Artin-Hasse Exponential, which is rather hard to understand and to use. The upshot, though, is that in the general case, you get
$$
1+\mathfrak m\cong\prod_{p\nmid m}W_\infty(\Bbb F_q)\,,
$$
those $W\!$s being the Witt rings, each being isomorphic to the integers in the unramified extension of $\Bbb Q_p$ of degree $[\Bbb F_q:\Bbb F_p]$.
