Trying to understand a proof of Fleck congruence I am trying to understand the proof of Fleck congruence given in Binomial coefficients modulo prime powers,  by A. Granville.
Fleck congruence states that for a prime $p$, integers $n \ge p$ and $0 \le r \le p-1$ and $q=\lfloor \frac{n-1}{p-1}\rfloor$
$$ \sum_{k \equiv r \bmod p}(-1)^k{n \choose k} \equiv 0 \pmod {p^q}.$$ 
The proof therein is very short and lean but it makes use of algebraic number theory, where my knowledge is very limited.
$\zeta$ being a $p$-th primitive root of $1$, I understand that 
$$ \sum_{k \equiv r \bmod p}(-1)^k{n \choose k} = \frac{1}{p}\sum_{i=0}^{p-1}\zeta^{-ir}(1-\zeta^i)^n .$$
I also understand that $(1-\zeta^i)^n$ belongs to the ideal generated by $(1-\zeta)^n$, since $\frac{1- \zeta^i}{1-\zeta}=(1+\zeta+\cdot\cdot+\zeta^{i-1})$ which is a unit (inversible) in the ring $\mathbb{Z}[\zeta]$,  for $1\le i \le p-1$.
It is also clear that $(1-\zeta)^{p-1}=p$. Then I would expect the quotient of $n$ by $p-1$ be involved but I fail to see why the quotient of $n-1$ by $p-1$ is eventually obtained, instead.  
Thanks for any clarification.
 A: We have ($i=0$ can be dropped)
$$
p\sum_{k\equiv r \ \mathrm{mod} \ p}(-1)^k\binom nk=(1-\zeta)^n\sum_{i=1}^{p-1} \zeta^{-ir}(1+\zeta+\cdots + \zeta^{i-1})^n. \ \ \ \ (1)
$$
Writing $\zeta=\zeta-1+1$ and $\zeta^{-ir}=\zeta^{pM-ir}$ for some $M$ with $pM>ir$, we see that 
$$
\sum_{i=1}^{p-1}\zeta^{-ir}(1+\zeta+\cdots+\zeta^{i-1})^n \equiv \sum_{i=1}^{p-1} i^n \ \mathrm{mod} \ (1-\zeta).
$$
The sum on the right is $0$ mod $p$ if $p-1\nmid n$, and $-1$ mod $p$ if $p-1|n$. 
Letting $\sum_{k\equiv r \ \mathrm{mod} \ p}(-1)^k\binom nk=X$ and taking norm to (1), we see that 
$$
p^{p-1} X^{p-1} = p^{n+d}K, \ \ \ (2)
$$
where $K\in\mathbb{Z}$, and  $\begin{cases} d\geq 1 &\mbox{if } p-1\nmid n \\ d=0  &\mbox{if } p-1|n\end{cases}$. 
If $p-1|n$, we have $\frac n{p-1}-1=\left\lfloor \frac{n-1}{p-1} \right\rfloor$
Thus, $\nu_p(X)\geq  \left\lfloor \frac{n-1}{p-1} \right\rfloor$.
If $p-1\nmid n$, write (2) in the form of 
$$
p^{p-1}X^{p-1}=p^{n+d'}K'
$$
where $d'\geq 1$ and $(K',p)=1$. 
Considering prime factorization of $X$, we obtain
$$
p-1+(p-1)\nu_p(X)=n+d',
$$
and hence $p-1|n+d'$.
Then $\nu_p(X)=\frac{n+d'}{p-1}-1\geq\left\lfloor \frac{n-1}{p-1} \right\rfloor $.
