Number theory question involving Wilson's theorem. How to solve?? Given p is a prime number greater than 2, and
$ 1 + \frac{1}{2} + \frac{1}{3} + ... \frac{1}{p-1} = \frac{N}{p-1}$ 
how do I show, $ p | N  $ ???
The previous part of this question had me factor $ x^{p-1} -1$ mod $p$. Which I think is just plainly $(x-1) ... (x-(p-1))$ 

Working $\bmod{p}$:
$$
1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{p-1}\\
\equiv 1 + 2^{-1} + 3^{-1} + \cdots + (p-1)^{-1}\\
\equiv 1 + 2 + \cdots + (p-1) \ \text{ reordered}
$$
which you have shown is $0\pmod{p}$ when you factored $x^{p-1} -1\pmod{p}$.
So if $\displaystyle 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{p-1} =\frac{N}{p-1}$ then necessarily $N(p-1)^{-1} = 0 \pmod{p}$ so $p$ divides $N$.
