Vtyjkyuk

Let $p$ be an odd prime and let $1leq n<p-1.$ Show that $$sum_t=1^pt^n equiv 0 pmodp$$

Remark: It seems one can not apply Fermat's little theorem directly as $n<p-1$

We might as well sum to $p-1$. Let $g$ be a primitive root of $p$. Then $1$, $2$, $3$, and so on up to $p-1$ are congruent, in some order, to $g^0$, $g^1$, up to $g^p-2$.
So modulo $p$ the terms in our sum are congruent to
$g^0$, $g^n$, $g^2n$, up to $g^(p-2)n$. Sum this geometric progression. We get $dfracg^(p-1)n-1g^n-1$. The numerator is congruent to $0$. There is no problem with the denominator, since $nlt p-1$.

If one prefers not to use fractions, equivalently note that
$$left(1+g^n+g^2n+cdots+g^(p-2)nright)(1-g^n)=1-g^(p-1)n.$$
The right-hand side is congruent to $0$. But $1-g^n$ is not congruent to $0$, and the result follows.

Since the sets of reduced residue classes $1, 2, . . . , (p − 1)$ and $g, 2g, . . . , (p − 1)g$ are the same if $(g,p)=1$ the reason being:
if $r_1gequiv r_2gpmod p,$ where $1le r_1< r_2le p-1$
$implies r_1equiv r_2$ which is impossible, so $r_1g≢r_2gpmod p$.

So, $sum_1le xle p-1 x^n equiv sum_1le yle p-1 (gy)^n pmod p$

$implies pmid(g^n-1)sum_1le xle p-1 x^n $

If we take $g$ to be a primitive root, $pmid (g^n-1)iff phi(p)=(p-1)mid n$

Since $1le n<p-1,p$ can not divide $(g^n-1)implies sum_1le xle p-1 x^nequiv 0pmod p$

If $(p-1)mid n, sum_1le xle p-1 x^nequiv sum_1le xle p-11pmod p=p-1equiv -1pmod p $

Call
$$ p_n = sum_k=1^p k^n. $$
Since, by Fermat's little theorem, $p_nequiv p_n-p+1pmodp$ for all $n > p-1$, it is sufficient to show that
$$ forall nin[0,p-2],quad p_nequiv 0pmodp, $$
where the cases $n=0$ and $n=1$ are trivial. Since:
$$ x(x-1)cdotldotscdot(x-p+1)equiv x^p-xpmodp, $$
if $e_k(x_1,ldots,x_p)$ is the $k$-th elementary symmetric polynomial, we have:
$$ forall kin[0,p-2],quad e_k(1,ldots,p)equiv 0pmodp, $$
so, in virtue of Newton-Girard identities and induction we have:
$$forall kin[0,p-2],quad p_kequiv 0pmodp, $$
QED.

EDIT:This is a wrong answer($r^n equiv t^n pmod p not Rightarrow r equiv t pmod p$). See the comments.

For $n=1$ is correct.

Now for any $n in 1,2,ldots,p-2$ we have that $1,2,ldots,p-1equiv 1,2^n,ldots,(p-1)^n pmod p$ since $r^n equiv t^n pmod p iff r equiv t pmod p$. Hence $$sum_t=1^pt^n equiv sum_t=1^pt equiv 0 pmodp .$$

WLOG assume $nk = p-1$ (if $n$ had factors coprime to $p-1$ they would just permute the sum, not change it's value, so remove them).

Put $S = (mathbb Z/pmathbb Z)^times$ and let $S^n$ denote the subgroup $ s in S $. The kernel of the map $S to S^n$ is the set of solutions of $z^n=1$ so $n|S^n| = |S|$. Details about subgroup here

This tells us that (mod p) the sum is equal to $n$ times the sum of the elements of $S^n$ so we just need to show that $$sum_s in S^ns equiv 0 pmod p$$ but since this is the set of solutions of $z^n-1=0$ by Vieta's formula they sum to $0$.