(This is a detail in my attempted answer of this MSE question) We look at $$f_n(x) = \sum_{k=1}^{n-1} (x+k)^n $$
I came to the following observation - for odd $n$ at the moment -, but do not see how prove it .
With the canonical primefactorization $n = p_1^{a_1}p_2^{a_2}...p_m^{a_m} $ we have also $w_n = p_1 p_2 ... p_m $ (this is sometimes denoted as $ rad(n)$), and I observed that, for odd n $$ f_n(x) \equiv 0 \pmod n \Leftrightarrow x = j \cdot w \qquad \qquad j \in \mathbb N$$
Q1: how can I prove, that if $x$ has the form, then the equivalence must be true? (might be easier than I think at the moment)
Q2: how can I proceed to prove, that also only if $x$ has that form the equivalence can be true? (seems to be much more difficult)
Hmm, Q1 seems to be not so difficult. We assume a simple case $n=pq^2$ being odd and $x=pq$ Then, modulo the odd prime $p$ we have $\pmod p$ $$\begin{array} {rcl} &&(pq+1)^{pq^2} + (pq+2)^{pq^2} + ... + (pq+(pq^2-1))^{pq^2} \\ &\equiv& (1)^{pq^2} + (2)^{pq^2} + ... + (-2)^{pq^2}+ (-1)^{pq^2} \\ &\equiv& (1)^{pq^2}+(-1)^{pq^2} + (2)^{pq^2} + (-2)^{pq^2} +... + ((n-1)/2)^{pq^2} +(-(n-1)/2)^{pq^2}\\ &\equiv& 0 + 0 + 0 + ... + 0 \\ &\equiv &0 \\&&\pmod p \end{array} $$
We can observe even more: for each primefactor $p_k$ of $n$ the above sum can be written as repeating partial sums of length $p_k$ such that we have $$\begin{array} {rcl} &&(pq+1)^n + (pq+2)^n + ... + (pq+n-1))^n \\ &\equiv& (1)^n + (2)^n + ... + (p-2)^n+ (p-1)^n+ (p)^n \\ && + (p+1)^n+(p+2)^n ...+ (2p-2)^n +(2p-1)^n + (2p)^n\\ && + (2p+1)^n+(2p+2)^n ...+ (3p-2)^n +(3p-1)^n + (3p)^n\\ &&...\\ &\equiv& 0 + 0 + 0 + ... + 0 \qquad \text{ (each row is zero by the same argument as above)}\\ &\equiv &0 \\&&\pmod p \end{array} $$ This is the same for all primefactors of $n$, and even if $n$ has more primefactors, so a zero occurs at the lowest common multiple of the primefactors which is just their product $pq = w_n$.
If $x=pq$ (or generalized the $rad(n)$) then the computation is the same for each primefactor of $n$ and thus the first zero accurs where the lengthes of the partial sums by each primefactor coincide and that is just the least common multiple of all the primefactors which is also their simple product and this is just $w_n$.
So we can see, that if $x=w_n$ or a multiple of $w_n$ then the partial sum from $x$ up to $x+w_n$ sums to zero modulo n