Take $\rm n=(p-1)a,a\in \mathbb{N}$.If $\rm i\in [1,p+1]-p $ then $\rm i^{(p-1)a}\equiv (i^{p-1})^a\equiv 1\pmod p$ and $\rm p^n\equiv 0\pmod p$. So $\rm \sum_{i=1}^{p+1}\equiv 1+1+...+0+1\equiv p\equiv 0\pmod p$.

After reading the problem statement, I somehow remembered this lemma: For primes \(p\) and integers \(k\),\[1^k+2^k+\cdots+(p-1)^k\equiv\begin{cases}-1&\text{if }p-1\mid k\\0&\text{otherwise.}\end{cases}\]Anyways take $n=(p-1)\alpha, \alpha \in N.$ We can even show there exist infinitely many positive integers $n$ such that $1^n + 2^n +... + (p + 1)^n \equiv 1\mod p$