TITLE: $\sum_{\zeta^p=1}(\zeta-1)^n$
QUESTION [8 upvotes]: 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 $0$?

REPLY [3 votes]: 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 $p=2$.