Let p be an odd prime number. Prove that the numerator of 1+1/2+1/3+...+1/(p-1) (when expressed as a single fraction with denominator (p-1)!) is divisible by p.
Let $\displaystyle N$ be numerator.
Then, $\displaystyle N\equiv \sum_{k=1}^{p-1} p! (k)^{-1} (\bmod p)$. Where $\displaystyle (k)^{-1}$ is inverse mod $\displaystyle p$.
But $\displaystyle \sum_{k=1}^{p-1} p! (k)^{-1} = p! \sum_{k=1}^{p-1}k$.
Since $\displaystyle (k)^{-1}$ is a permuation of $\displaystyle \{1,2,...,p-1\}$.
Finally, $\displaystyle \sum_{k=1}^{p-1} k = \frac{p(p-1)}{2}$.
Therefore, it is divisible by $\displaystyle p$.