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For a prime $p$, $p$ divides $\sum_{i=1}^{p-1}i^k$ if and only if $p-1$ does not divide $k$. Thus for two primes you may simply suppose that $p_1<p_2$, then $p_1p_2-1\equiv p_1-1 \pmod{p_2-1}$ is not divisible by $p_2-1$.