Power and congruence

[Thanic Nur Samin]

Let $p$ be an odd prime.Prove that,$1^i+2^i+...(p-1)^i\equiv0 \pmod p$ For every $i$ from $0$ to $p-2$

[mutasimmim]

What about $p=3$ and $i=0$ ?

[Phlembac Adib Hasan]

What about $p=3$ and $i=0$ ?

I think he did a typo. The bound should be $1\leq i\leq p- 2$.

There is a easy solution by induction. Define \[S_i = \sum^{p-1}_{j=1}j^i\] Firstly note $p|S_1= p\cdot \frac {p-1}2$. Now prove that for $i\ge 2$, we have
\[p^{i+1}-p=\binom{i+1}iS_i+\sum ^{i-1}_{j=1}\binom {i+1} j S_j\]
From inductive hypothesis, $p$ divides every $S_j$. So it follows $p|S_i$ since $p\not | \binom{i+1}i$