Proof that $\sum\limits_i \frac{(p-1)!}i$ is divisible by $p$ 
Problem: Let $p$ be a prime. Consider $\sum\limits_{i = 1}^{p-1} \frac1i = \frac K{(p-1)!}$. Rearranging, we have
  $K = \sum\limits_{i = 1}^{p-1} \frac{(p-1)!}i$.
  Prove that $p \mid K$. 

Hint: consider the factorization $x^{p-1} - 1 \equiv (x-1) ...(x - (p-1)) \pmod{p}$.
Attempt: I am finding applying the hint difficult. Sure, the RHS of the congruence relation contains $(p-1)!$, but I can't see how the factorization would help me prove the expression for $K$. Any help is appreciated.
 A: This is not a complete proof but rather some thoughts.
Edit: I erroneously states that this works for p=2 when it does not. 
Suppose $p$ is an odd prime. By the hint, we have that $$p^{p-1} - 1 \equiv (p-1)! \pmod p.$$ On the other hand $$p^{p-1} -1 \equiv -1 \pmod p.$$ Putting these two facts together gives $$(p-1)! \equiv -1 \pmod p.$$ Therefore, $$K \equiv \sum_i^{p-1}\frac{-1}{i} \pmod p.$$ Can you finish from here? 
A: I have a much simpler solution. But I will start by stating that this is problem number 60 from this book (more complicated since it asks for $p^2 \mid K, p>3$) and the solution below is not from the book. So, for $p>2$ we have $p-1$ even and
$$\sum\limits_{i=1}^{p-1}\frac{1}{i}=\sum\limits_{i=1}^{\frac{p-1}{2}}\frac{1}{i}+\sum\limits_{i=\frac{p+1}{2}}^{p-1}\frac{1}{i}=
\sum\limits_{i=1}^{\frac{p-1}{2}}\frac{1}{i}+\sum\limits_{i=1}^{\frac{p-1}{2}}\frac{1}{p-i}=\\
\sum\limits_{i=1}^{\frac{p-1}{2}}\left(\frac{1}{i}+\frac{1}{p-i}\right)=
\sum\limits_{i=1}^{\frac{p-1}{2}}\frac{p}{i(p-i)}=p\sum\limits_{i=1}^{\frac{p-1}{2}}\frac{1}{i(p-i)}=p\frac{A}{(p-1)!}$$
where $K=pA$ and $A \in \mathbb{N}$.
