How to prove that the $p$-adic units can be written as $$\mathbb{Z}_p^\times \cong \mu_{p-1}\times(1 + p\mathbb{Z}_p) \cong \mathbb{Z}/(p-1)\mathbb{Z}\times\mathbb{Z}_p$$ where $\mu_n$ is the $n$-th roots of unity in $\mathbb{Z}_p$?
Here $p>2$ is a prime number.
Any hint or link would be helpful.
Besides, it looks strange: why the units of $\mathbb{Z}_p$ is isomorphic to some guy which looks bigger than $\mathbb{Z}_p$?
This must be covered in almost every text on the $p$-adic numbers; I think the book of Gouvêa is the best of these.
And your statement is not quite true: for $p=2$, $\mu_{p-1}$ is trivial all right, but the other part is isomorphic to $\{\pm1\}\times(1+4\Bbb Z_2)$. My tale below omits the story for $p=2$, and you can fill this in yourself.
First, you can consider the units, $\Bbb Z_p^\times$, and reduce them modulo $p$ to the multiplicative group of $\Bbb F_p\cong\Bbb Z/p\Bbb Z$. It’s cyclic of order $p-1$, as I’m sure you know. So we have an exact sequence: $$ 0\longrightarrow K\longrightarrow\Bbb Z_p^\times\longrightarrow\Bbb F_p^\times\longrightarrow0\,; $$ if you’re unfamiliar with the notation of exact sequences this merely says that there’s a surjective map from the middle term to the one to its right, with kernel equal to the one to its left.
What’s the kernel? It’s the units that go to $1$ in the field $\Bbb F_p$, in other words $1+p\Bbb Z_p$. To prove that $\Bbb Z_p^\times$ is the direct product of the two things to either side of it, it’s enough to show that there’s a homomorphism from $\Bbb F_p^\times$ into it whose image hits $K$ only in the identity. This is the fun part:
You can find $(p-1)$-th roots of unity in $\Bbb Z_p$ either by a routine application of any version of Hensel’s Lemma that you like, or, my favorite method, take an element of $\Bbb F_p^\times$, lift it to any element of $\Bbb Z_p$ that goes to it modulo $p$, and take successive $p$-th powers: $x\mapsto x^p\mapsto(x^p)^p\mapsto\cdots$ etc. I’ll leave it to you to show that this is a good convergent sequence, and its limit is clearly a suitable root of unity.
Showing that the multiplicative group $1+p\Bbb Z_p$ is isomorphic to the additive group $\Bbb Z_p^+$ is rather less fun. To tell you the truth, when I first saw the logarithmic argument, I didn’t like it, but I now think it’s the best one. You have to convince yourself that as long as the series for $\log(1+x)$ that you saw in Calculus is convergent, then $\log\bigl[(1+x)(1+y)\bigr]=\log(1+x)+\log(1+y)$, and for $x\in p\Bbb Z_p$, the series is convergent. And the values of the log are all in $p\Bbb Z_p\cong\Bbb Z_p$ (as additive groups, of course), and fill out that group.
Here is an alternative approach using some category theory.
From the universal property of the inverse limit we have that the inverse limit functor is right-adjoint. The group of units functor is right adjoint to the integral group ring functor. Thus both of these functors preserve limits.
We know from group theory that for $p$ an odd prime, $(\mathbb{Z}/p^n\mathbb{Z})^\times$ is cyclic of order $p^n-p^{n-1}=p^{n-1}(p-1)$, thus by the Chinese Remainder Theorem $(\mathbb{Z}/p^n\mathbb{Z})^\times\cong \mathbb{Z}/p^{n-1}\mathbb{Z}\times\mathbb{Z}/(p-1)\mathbb{Z}\cong \mathbb{Z}/p^{n-1}\mathbb{Z}\times (\mathbb{Z}/p\mathbb{Z})^\times$.
Putting these two statements together: $$ (\mathbb{Z}_p)^\times\cong(\lim\limits_{\longleftarrow}\mathbb{Z}/ p^n\mathbb{Z})^\times\cong\lim\limits_{\longleftarrow}(\mathbb{Z}/p^n\mathbb{Z})^\times\cong \lim\limits_{\longleftarrow}(\mathbb{Z}/p^{n-1}\mathbb{Z}\times (\mathbb{Z}/p\mathbb{Z})^\times)$$
$$ \cong \lim\limits_{\longleftarrow}\mathbb{Z}/p^{n-1}\mathbb{Z}\times\lim\limits_{\longleftarrow} (\mathbb{Z}/p\mathbb{Z})^\times\cong \mathbb{Z}_p\times (\mathbb{Z}/p\mathbb{Z})^\times $$
