Prove $1^r+2^r+.....\equiv 0\pmod p$ , given that p is odd prime and $0<r<p-1$

There is another way to do this, I am not sure if this type of approach is what you had in mind:

Claim 1: Let $H$ be a subgroup [under multiplication] of $(\mathbb{F}_p)^{\times}$ that satisfies $|H| \ge 2$. Then $\sum_{a \in H} a \equiv_p 0$.

Indeed, we first note the following: For any $a_0 \in H$ the sets $\{a; a \in H\}$ and $\{a_0a; a \in H\}$ are the same. Thus for any $a_0 \in H; a_0 \not = 1$; we note the following:

$$a_0\sum_{a \in H} a = \sum_{a \in H} a_0a \equiv_p \sum_{a' \in H} a' =\sum_{a \in H} a.$$

However, as $a_0 \not =1$ the above implies that $\sum_{a\in H} a$ must be 0 mod $p$ and so Claim 1 follows. $\surd$

Now for each positive integer $r$, the set of $r$-th powers of $(\mathbb{F}_p)^{\times}$ form a subgroup $H_r$ of $(\mathbb{F}_p)^{\times}$, and for all $a \in H_r$ the number of solutions to $x^r=a$ is the same $\frac{p-1}{|H_r|} < p-1$ for all $r<p-1$ [because $(\mathbb{F}_p)^{\times}$ is cyclic and so has an element of order $p-1$]. Thus $|H_r| >1$ and so

$$\sum_{x \in \mathbb{F}_p} x^r = 0^r + \frac{p-1}{|H_r|} \sum_{a \in H_r} a, $$

is indeed divisible by $p$ by Claim 1.