I wish to ask today's Putnam problem B6:
Suppose $p$ is an odd prime. Prove that for $n\in \{0,1,2...p-1\}$, at least $\frac{p+1}{2}$ number of $\sum^{p-1}_{k=0} k! n^{k}$ is not divisble by $p$.
For example, for $p=3$, we have $(0,1,2)$ corresponds to $(0, 1+1+2,1+2+2*4)$. For $p=5$, we have $(0,1,2,3,4)$ corresponds to $(0,1+1+2+6+24,1+2+2*4+6*8+24*16..)$.
It is to be expected that an official solution can be found in somewhere (the Putnam archive, American Mathematical Monthly, AOPS, etc). My point of raising this question is not to ask a solution of it, because I am not interested in problem solving techniques. Instead I want to ask the following questions:
1) Is the function $\sum\limits_{k=0} ^{p-1} k! n^{k}$ well known? Can it be expressed in some closed form via generating functions or other combinatorical tools? My combinatorical background is very weak so I think I should ask others.
2) In general how strong the statement is? For example, for $p$ quite large, how many elements in $\mathbb{Z}_{p}$ tend to satisfy $\sum \limits_{k=0} ^{p-1} k! n^{k}=0$?
3) It is possible to see the problem via some 'high level' proof? (I know David Speyer is found of this kind of thing). Because I am not specialized in number theory, the limited number theory knowledge I have in Dedekind domains and valuations seems to be quite irrelevant to this problem. But I think there should be some way this small problem generalizes to something more interesting to an expert.