$\sum_{\zeta^p=1}(\zeta-1)^n$ Given $n\geq0$ let
$$
z_n=\sum_{\zeta^p=1}(\zeta-1)^n
$$
where $p$ is an odd prime number (summation extended to all $p$-th roots of 1). It is clear that:


*

*$z_n\in\Bbb Z$ (it's a Galois invariant sum of integral elements in the extension $\Bbb Q(\zeta)$).

*$p^m\mid z_n$, where $p^m$ is the maximal power of $p$ in $\frak p^n$ where $\frak p$ is the unique prime in $\Bbb Z[\zeta]$ over $p$ (see Jyrki Lahtonen's comments below), i.e. $m=\lceil\frac n{p-1}\rceil$).
It is an obvious computation that 
$$z_n=\sum_{\zeta^p=1}\sum_{k=0}^n\binom nk\zeta^k(-1)^{n-k}=\sum_{k=0}^n\binom nk(-1)^{n-k}\sum_{\zeta^p=1}\zeta^k
$$ 
and since $\sum_{\zeta^p=1}\zeta^k=0$ if $(p,k)=1$ and $=p$ if $p\mid k$ we eventually get
$$
z_n=(-1)^np\sum_{\ell=0}^{n/p}(-1)^\ell\binom n{\ell p}
$$
(clearly the summation stops at the greatest integer smaller than $n/p$ if $n/p\notin\Bbb Z$). Now my questions are:


*

*can we get a more compact expression for $z_n$ and/or some characterizing description?

*The divisibility mentioned above translates implies that $lim_{n\to}|z_n|_p=0$ ($p$-adic absolute value). Can we be more precise about the speed of convergence to $0$?
 A: I’m not sure that this is the sort of answer you’re hoping for, for the last question, about the speed of convergence to zero of your numbers, but let me put it forward.
The question is purely local, so we may as well consider it over the $p$-adic integers $\mathbb Z_p$. Your number $\zeta-1$ is a prime in the field $K$ of $p$-th roots of unity over $\mathbb Q_p$, so I’ll call it $\pi$. We have $v(\pi)=1/(p-1)$, where I’m using the (additive) valuation $v$ normalized so that $v(p)=1$. We may consider, for each $n$, the $\mathbb Z_p$-polynomial $f_n(X)=\prod_\sigma(X-\sigma\pi^n)$, the product being taken over the $p-1$ elements of the Galois group. This polynomial, if irreducible, is the minimal polynomial for $\pi^n$, though I don’t make any claim of irreducibility. What’s important is that (1) your numbers $z_n$ are the coefficients of $X^{p-2}$, up to sign; and (2) the Newton polygon of $f_n$ consists of a single segment, running from $(0,n)$ to $(p-1,0)$.
Thus $v(z_n)\ge n/(p-1)$.
A: Many of the results can be easily seen through using a Mahler series expansion as a generating function. Specifically, we have the continuous function $\chi_{p^t\mathbb{Z}_p} :\mathbb{Z}_p \to \{0,1\}$, which is the locally constant indicator function,
$$\chi_{p^t\mathbb{Z}_p}(x)  = \begin{cases} 1 & x \in p^t\mathbb{Z}_p \\  0 & x \not\in p^t\mathbb{Z}_p\end{cases}$$
Clearly, its Mahler series has coefficients $a_n = \nabla^n \chi_{p^t\mathbb{Z}_p}(0)$ in $\mathbb{Z}$,
$$\chi_{p^t\mathbb{Z}_p}(x)=\sum_{n\ge 0} a_n \binom{x}{n}$$
However, notice that we have the alternative representation in $\mathbb{Q}_p(\zeta_{p^t})$,
$$\chi_{p^t\mathbb{Z}_p}(x) = \frac{1}{p^t}\sum_{k=0}^{p^t-1} \zeta_{p^t}^{kx}$$
We can now evaluate its finite differences at $x=0$,
$$\nabla^n \chi_{p^t\mathbb{Z}_p}(0) = \frac{1}{p^t}\sum_{k=0}^{p^t-1}(\zeta_{p^t}^k-1)^n  = \frac{z_n}{p^t}$$
At this point, the original question is concerned with the $t=1$ case, but I'll keep the argument general. Because Mahler series representations are unique, the integer (in $\mathbb{Z}$) $a_n = \frac{z_n}{p^t}$ already shows us that $z_n$ is an integer divisible by $p^t$. Furthermore, Mahler series have coefficients that satisfy $\lim_{n \to \infty} |a_n|_p = 0$, so $\lim_{n \to \infty} |z_n|_p\ = 0$.
We can also be more precise on the speed of convergence since $\chi_{p\mathbb{Z}_p}$ is strictly differentiable. It implies that $\lim_{n \to \infty} n|a_n|_p = 0 $ so we have,
$$\lim_{n \to \infty} n|z_n|_p = 0 $$

While lying in bed this morning I realized we also get the same formula from the question as well just by the formula for the Mahler coefficients, the characteristic function simply screens out anything that is not a multiple of $p^t$ for us.
$$a_n = \sum_{k=0}^n \binom{n}{k} (-1)^{n-k} \chi_{p^t\mathbb{Z}_p}(k)= \sum_{k=0}^{ \lfloor n/p^t \rfloor} \binom{n}{p^tk} (-1)^{n-p^tk} $$
$$z_n = (-1)^n p^t \sum_{k=0}^{ \lfloor n/p^t \rfloor} \binom{n}{p^tk} (-1)^{pk} $$
The $(-1)^{pk}$ is left to account for the case when $p=2$.
