Prove that for every p - prime, $p>3$ the numerator of sum of $p-1$ summands of harmonic series is divisible by $p$ square.
How prove this $(p-1)!\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{p-1}\right)\equiv 0\pmod{p^2}$