Newton's Binomial Series and p-adic roots of unity I've been playing around with $p$
 -adic exponentiation via Newton's binomial series: $$\left(1+t\right)^{x}=\sum_{k=0}^{\infty}t^{k}\binom{x}{k},\textrm{ }x\in\mathbb{Z}_{p}$$
 where $t\in\mathbb{C}_{p}$
  with $\left|t\right|_{p}<1$
 . I should mention right away that I am most concerned with the case where $p=3$
 , but answers for the case of an arbitrary prime $p$
  would also be much appreciated. Throughout this, $x$
  is an indeterminate/variable in $\mathbb{Z}_{p}$.
Now, for any integer $a\geq2$, let $\xi_{a}$ denote a primitive $a$th root of unity. Moreover, let's assume that we have chosen a specific embedding/labeling of $\xi_{a}$ in $\mathbb{C}_{p}$, so that $\xi_{a}$ denotes a single element of $\mathbb{C}_{p}$. 
My hope is to try to use the Newton series formula to make sense of the function $x\in\mathbb{Z}_{p}\mapsto\xi_{a}^{x}\in\mathbb{C}_{p}$. As an example, for $p=3$, note that:$$\frac{x^{3}-1}{x-1}=x^{2}+x+1=\left(x-\xi_{3}\right)\left(x-\xi_{3}^{2}\right)$$
 Setting $x=1$, I obtain:$$3=\left(1-\xi_{3}\right)\left(1-\xi_{3}^{2}\right)=\frac{1}{\xi_{3}}\left(1-\xi_{3}\right)\left(\xi_{3}-1\right)$$
 and so:$$\frac{1}{3}=\left|-3\xi_{3}\right|_{3}=\left|\xi_{3}-1\right|_{3}^{2}$$
 from which I deduce that: $$\left|\xi_{3}-1\right|_{3}=\frac{1}{\sqrt{3}}<1$$
 and hence, that: $$\xi_{3}^{x}=\left(1+\xi_{3}-1\right)^{x}=\sum_{k=0}^{\infty}\left(\xi_{3}-1\right)^{k}\binom{x}{k}$$
 is a continuous, (strictly) differentiable (and analytic?) function from $\mathbb{Z}_{3}\rightarrow\mathbb{C}_{3}$
 . Obviously, this construction depends on both the values of $a$
  and $p$. 
This leads me to my questions:
1) there a general, closed-form expression for $\left|\xi_{a}-1\right|_{p}$? For $\left|\xi_{a}-1\right|_{3}$? Equivalently: given $p$
 , for what $a\geq2$
  does: $$\xi_{a}^{x}=\left(1+\xi_{a}-1\right)^{x}=\sum_{k=0}^{\infty}\left(\xi_{a}-1\right)^{k}\binom{x}{k}$$
  define a strictly differentiable (analytic?) function from $\mathbb{Z}_{p}$
  to $\mathbb{C}_{p}$?
2) On a related note, if $a$
  is even, is $\xi_{a}^{a/2}=-1$
  in $\mathbb{C}_{p}$? That is to say, when $a$
  is even, does there exist a choice of a specific embedding of $\xi_{a}$
  into $\mathbb{C}_{p}$
  so that $\xi_{a}$
  satisfies $\xi_{a}^{a/2}=-1$?
 A: This is something I actually know something about.
I think I’m going to make things hard for you and insist on the use of the additive valuation $v_p$ instead of the multiplicative absolute value $\vert\star\vert_p$. Remember that $\vert z\vert_p=p^{-v_p(z)}$. So a number $z$ is small if its “order” $v_p(z)$ is big. And remember that $\vert z\vert=1$ if and only if $v_pz=0$.
Although the absolute-value functions are normalized in various ways dependent on the context, the order function $v_p$ is almost always normalized so that $v_pp=1$. So $v_p(\sqrt[n]p\,)=1/n$.
If $k\supset\Bbb Q_p$ is a finite field extension, then $v_p$ extends uniquely by the rule
$$
v_p(z)=\frac{v_p\bigl(\text N^k_{\Bbb Q_p}(z)\bigr)}{[k:\Bbb Q_p]}\,,
$$
where $\text N^k_{\Bbb Q_p}$ is the field-theoretic norm from $k$ down to $\Bbb Q_p$. Note that if $[k:\Bbb Q_p]=n$, the norm of an element $z\in\Bbb Q_p$ is $z^n$, so the formula does really define an extension of the order-function to $k$. You can use strong Hensel to show that the newly-defined $v_p$ still satisfies the necessary relation $v_p(z+w)\ge\min(v_pz,v_pw)$.
Now to your questions, which are perhaps somewhat more interesting for larger primes than $3$. I’ll use your notation $\zeta_n$ for a primitive $n$-th root of unity in an extension of $\Bbb Q_p$, and tell you the results:
If $n$ is prime to $p$, then $v_p(\zeta_n-1)=0$: in other words, these roots of unity are maximally far from $1$.
For $n=p^m$, we get $v_p(\zeta_{p^m}-1)=\frac1{p^{m-1}(p-1)}$, and now you see why I insisted on $v_p$ instead of the absolute-value notation. In particular, this formula confirms your conclusion that $v_3(\zeta_3-1)=1/2$; and it agrees with the observation that $v_2(\zeta_2-1)=1$, after all we’re talking about $-2$ there.
I’ll leave it to you to see what $v_p(\zeta_n-1)$ is when $n$ is neither a power of $p$ nor prime to $p$.
Now to your question about the Binomial Series. You seem to recognize that the binomial coeffients $\binom xm$ are $p$-adic integers whenever $x$ is a $p$-adic integer, and that this means convergence of the Binomial series for $(1+z)^x$ whenever $v_p(z)>0$. Even when $z$ is all the way up in $\Bbb C_p$. (I like to point out that you may plug elements $z$ of the algebraic closure of $\Bbb Q_p$, even though this field is not complete: after all, $z$ will be in a finite extension of $\Bbb Q_p$, good enough for convergence.)
My opinion, however, is that you made a bad, or at least uninteresting, choice of elements $z$ to plug into the series for $(1+z)^x$. For, look at this: write your $x$ as $a_0+a_1p+a_2p^2+\cdots+a_np^n+\cdots$. When you substitute $\zeta^{p^n}-1$ for $z$ in the formula, you get $\zeta_{p^n}^\lambda$, where $\lambda$ is the finite sum $\sum_{i=1}^{n-1}a_ip^i$. In other words, the final tail of $x$ doesn’t matter: your function is locally constant, and certainly differentiable, analytic, etc. Much more interesting than your choice would be an element of $\Bbb C_p$ that isn’t a root of unity. Then all should be well, you should get an analytic function of $x$ then.
As a closing note, let me point out that if your picture of roots of unity in the $p$-adic world is formed from the picture in the complex world, you will be led badly astray. The complex roots of unity are dense in the unit circle, but the $p$-adic roots of unity are unfriendly to each other: if $\zeta$ and $\zeta'$ are two such, then $v_p(\zeta-\zeta')\le\frac1{p-1}$.
