Prove that if $p\ge 7$ then $\exists n\in\Bbb{Z}$ such that $10^{n(p-1)}\equiv1 \mod 9p$. Prove that if $p\ge 7$ then $\exists n\in\Bbb{Z}$ such that $10^{n(p-1)}\equiv1 \mod 9p$. Edit: $p$ is prime, of course.
I tried using theorems regarding Euler, but I can't seem to arrive at something useful. I could really use your guidance. 

By pigeon-hole, there exist $k,m$ with $10^{k(p-1)}\equiv 10^{m(p-1)}$ and wlog. $k>m$. Since $10$ is coprime to $9p$, we can cancel $10^{m(p-1)]}$ and find $10^{(k-m)(p-1)}\equiv 1\pmod{9p}$.
