I suppose $(a,m)=(a-1,m)=1$, show that $1+a+a^2+\ldots+a^{\phi(m)-1}\equiv0\pmod m$

Question

I suppose $(a,m)=(a-1,m)=1$, show that $$1+a+a^2+\ldots+a^{\phi(m)-1}\equiv0\pmod m$$

I tried $$1+a+a^2+a^{\phi(m)-1}=\frac{a^{\phi(m)}-1}{a-1}$$ I believe that this equality that put up help, is it?

Answer 1

Notice how $$m|a^{\phi(m)} - 1$$ by Euler-Fermat Theorem. Since $(a-1,m)=1$, this means $$m|\frac{a^{\phi(m)}-1}{a-1} = 1+a+...+a^{\phi(m)-1}.$$