Show that $10^{n(p-1)}\equiv 1\pmod{\! 9p}$ for $p\ge 7$ I need to prove that for each prime $p \ge 7$ and for each $n \in\Bbb N$ 
$$10^{n(p-1)} \equiv 1 \pmod {9p}$$
What I've tried: 
I know $10$ is coprime to $9$ and $p$, so it is coprime to $9p$.
I know that $\,\varphi (9p) = \varphi (9)\cdot \varphi (p) = 6(p-1)$
and now I'm stuck. I guess I'm missing some basic rule.
Any help or hint will be appreciated.

By Fermat's little theorem, since $(10,p)=1$,
$$10^{n(p-1)}\equiv \left(10^{p-1}\right)^n\equiv 1^n\equiv 1\pmod{\! p}$$   
Also $$10^{n(p-1)}\equiv 1^{n(p-1)}\equiv 1\pmod{\! 9}$$   
Since $(9,p)=1$, we have $$9,p\mid 10^{n(p-1)}-1\iff 9p\mid 10^{n(p-1)}-1$$
