Asking 2011 Putnam B6

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.

Edit: Asked on Overflow: There are some very good answer which can be found here which explain some deeper meaning: https://mathoverflow.net/questions/82648/truncated-exponential-series-modulo-p-deeper-meaning-for-a-putnam-question

I feel that looking at $\sum_{k=0}^{p-1} k!n^k$ without changing its form is the wrong way to approach this problem, and if there are deeper solutions they will come by modifying this first. After some rearrangements modulo $p$, what we are really looking at is the truncated exponential series.

Rephrasing the question: First, let us ignore $n=0$, as in this case the sum is just $1$, and instead consider only the multiplicative group. Letting $k=p-1-l$ and using Wilsons theorem and Fermats little theorem we can rewrite our sum as $$\sum_{k=0}^{p-1}k!n^{k}\equiv\sum_{l=0}^{p-1}(p-1-l)!n^{p-1-l}\equiv-\sum_{l=0}^{p-1}\frac{1}{(p-l)(p-l+1)\cdots(p-1)}n^{-l} $$ $$\equiv -\sum_{l=0}^{p-1}(-1)^{l}\frac{n^{-l}}{l!}.$$ Using the fact that $n\rightarrow p-n $ and $n\rightarrow n^{-1}$ are isomorphisms of the multiplicative group, we have reduced the problem to considering for which $n$ we have $$-\sum_{l=0}^{p-1}\frac{x^{l}}{l!}=0.$$ This means that the original problem is equivalent to bounding the number of zeros modulo $p$ of the truncated exponential function $$E(z)=\sum_{l=0}^{p-1}\frac{z^{l}}{l!}.$$

Proof of the question: Consider $$Q(z)=z^{p}-z+\sum_{l=0}^{p-1}\frac{z^{l}}{l!}.$$ Then for each integer $Q(n)=E(n).$ However, $$Q^{'}(z)\equiv E^{'}(z)-1=E(z)-\frac{z^{p-1}}{(p-1)!}-1\equiv E(z)+z^{p-1}-1.$$ Then, if $Q(n)=0$ for $n\neq0$ , we must also have $Q^{'}(n)=0$ so that $n$ is a double root of $Q(n).$ Since $\deg Q(n)=p$, we see that at most half of the integers $p\in\left\{ 1,2,\dots,p-1\right\}$ satisfy $E(n)=0.$ Since $E(0)=1$, we conclude the desired result.

Motivation: The first idea I had looked at $E(z)-E(-z)$, but if you carry through with the computations you can only show that at least $\frac{p-1}{4}$ are non zero. Instead, lets just look at the derivative of $E(z)$ which is $E(z)+z^{p-1}$. Knowing that multiple zeros are what we want as it means less zeros total, we must add a function whose derivative is $-1$ since this will force multiple zeros. The obvious choice is then $x^p-x$, as it leaves the original polynomial invariant.

Remark: If you ask about the truncated exponential, there are many papers in existence and probably a high level way to see the problem. For example, see the first part of http://www.science.unitn.it/~mattarei/Ricerca/Preprints/AH.pdf