Proving that numerator of harmonic series is divisible by p Let p be a prime number. Prove that the numerator of $1+$$1 \over 2$+ $1 \over 3$+ $\cdots$  + $1 \over p-1$ (when expressed as a single fraction) is divisible by p.
I think this may have something to do with the fact that each term here has one inverse? Not sure how to express that in math terms though. 
 A: Note that $1+ \frac{1}{2} + \ldots + \frac{1}{p-1} = \left(1+ \frac{1}{p-1}\right) + \left(\frac{1}{2}+ \frac{1}{p-2}\right) + \ldots + \left(\frac{1}{\left\lfloor\frac{p}{2}\right\rfloor}+ \frac{1}{\left\lfloor\frac{p}{2}\right\rfloor+1}\right)$.
This evaluates to:
$$
\left(\frac{p}{p-1}\right) + \left(\frac{p}{2(p-2)}\right) + \ldots + \left(\frac{p}{\left(\left\lfloor\frac{p}{2}\right\rfloor \right)\left(\left\lfloor\frac{p}{2}\right\rfloor+1\right)}\right)
$$
The denominator is not divisible by $p$, because it is $(p-1)!$ which is not a multiple of $p$ as $p$  is prime. The numerator is clearly divisible by $p$. Hence the numerator of the reduced fraction is also divisible by $p$.
Read up Wolstemholme's theorem. It states that in fact, the numerator is divisible by $p^2$.
A: Well, writing the harmonic series as a single factor give
\begin{align}
\sum^{p-1}_{k=1} \frac{1}{k} = \sum^{p-1}_{k=1} \frac{1\cdot 2\cdots \hat k \cdots (p-1)}{(p-1)!} =:\sum^{p-1}_{k=1} \frac{a_k}{(p-1)!} \ \ \ \ (*)
\end{align}
where $\hat k$ indicates the missing multiple. Using Wilson's theorem which states
\begin{align}
ka_k=(p-1)! \equiv -1 \mod p \ \ \Rightarrow \ \ a_k \equiv -k^{-1} \mod p
\end{align}
which means
\begin{align}
\sum^{p-1}_{k=1} a_k \equiv -\sum^{p-1}_{k=1} k^{-1}  \mod p.
\end{align}
Essentially, we are summing over units in $\mathbb{Z}/p\mathbb{Z}$ which is zero since
\begin{align}
1+2+\ldots +(p-1) = \frac{p(p-1)}{2}.
\end{align}
Note: It should be noted that the denominator of $(*)$ is not divisible by $p$ which means it will not cancel out the $p$ in the numerator when you look at the reduced fraction of $(*)$. 
A: I believe the simplest approach is the following, also given by Jacky Chong:
Rewrite the terms from $1$ to $1/(1-p) $ with the denominator $ (1-p)! $, the kth term being 
$ (p-1)!/k(p-1)! $ 
If $ (p-1)!/a(p-1)! = (p-1)!/b(p-1)! $ then we must have a=b.  Therefore the denominators constitute $p-1$ different numbers (mod p) from $1$ to $p-1$, none of which is $0$ (mod p), and the same is true for the nominators.
Therefore the same is trues for the terms $ (p-1)!/k(p-1)! $. They constitute a permutation of the numbers 1 to p-1 (mod p). 
Therefore their sum is (from the arithmetic series formula) $p(p-1)/2$ (mod  p) , that is $= 0$ (mod p). It doesn't show that the sum is also $0$ (mod $p^2$)
