Let $p$ be an odd prime, let $r_1, r_2,..., r_{p-1}$ be the integers $1, 2,..., p-1$ in any order. Prove that at least two of the numbers $1\cdot r_1, 2\cdot r_2, ..., (p-1)\cdot r_{p-1}$ are congruent modulo $p$.
Here's what I have noticed so far. I think this is likely true by working with cases of $p=5$ and $p=7$. I also noticed that:
$$\begin{align}1\cdot r_1&=r_1 \pmod p\\2\cdot r_2&=2r_2 \pmod p\\... \\(p-2)\cdot r_{p-2}&=-2\cdot r_{p-2} \pmod p\\(p-1)\cdot r_{p-1}&=-r_{p-1} \pmod p\end{align}$$
I'm not sure where to go from here, or if I'm even on the right track. Would the best method be a direct proof or should I consider a proof by contradiction? Thanks for the help, and sorry for the inability to type subscripts in here better.