What does root of unity in $\mathbb{Z}_p$ look like? Let $p$ be an odd prime. Then by Hensel's lemma it's clear that $\mathbb{Z}_p $ contains all $(p-1)^{th}$ root of unity which reduces to $1$, $2$, ... , $p-1$ in $\mathbb{F}_p$. 
My question is do we know what they look like? What I mean is do we know what they look like in the form $\sum\limits_{n=1}^\infty {a_np^n}$ where $\{a_n\}$ is somewhat nice sequence.
Or do we not need to know? What I am really interested in is the galois extension $\mathbb{Q}_p\{\mu_{p^\infty}\} /\mathbb{Q}_p$ which has galois group $\mathbb{Z}_p^\times =\mu_{p-1} \times 1+ p\mathbb{Z}_p \cong \mathbb{Z}/(p-1)\mathbb{Z} \times \mathbb{Z}_p$ and the field fixed by $\mu_{p-1}$. And of course by $\mathbb{Q}_p\{\mu_{p^\infty}\}$ I mean the field obtained by adjoining all $p$-power roots of unity onto $\mathbb{Q}_p$.
It's first time posting a question here so let me know if I broke a rule or asked in a way it is really bad to understand. Thanks!
Edit $1$
I tried explicitly computing it for low $p$ and I do not see any nice patterns. Although this could be because I haven't iterated it too many times. Any software or program I could use to compute the roots of unity of this form easily? I am not good at programming in general though.
 A: No, they’re irrational, so there’s certainly no pattern in the digits of the standard $p$-ary expansion.
There are several ways to calculate the roots of unity explicitly. Here’s a brilliantly easy, but slow way: take your integer $n$ with $1<n<p-1$, and just successively raise it to the $p$-th power. That is, perform $z\mapsto z^p$ repeatedly, starting with $z=n$. You get one more digit per iteration.
One of the quickest ways is to use Newton-Raphson, which you must know from Calculus. You’re solving $X^{p-1}-1=0$, starting with your first guess $n$ with, as before, $1<n<p-1$. Your first error is $n^{p-1}-1$, and you divide this by the derivative of your function $(p-1)X^{p-2}$, evaluated at $n$. The quotient is your correction, and you subtract this from your first guess of $n$. That givew you your next approximation. I could write the formulas down, but they’re precisely what you read in your calculus text. Anyhow, you get twice as much accuracy in each iteration. I just did it for $i\in\mathbb Z_5$, and got 30 digits with 5 iterations. (Makes sense, since $2^5>30$.)
There’s a way of using the Binomial Expansion of $(1+X)^a$, and it can be used for finding $m$-th roots of other numbers than $1$, but it’s no faster-converging than the first method I suggested, and of course much more difficult computationally.
