I have been stuck on this exercise for far too long:
Show that if $a$ and $m$ are positive integers with $(a,b)=(a-1,m)=1$, then $1+a+a^2+\cdots+a^{\phi(m)-1}\equiv0\pmod m.$
First of all, I know that $1+a+a^2+\cdots+a^{\phi(m)-1}=\frac{a^{\phi(m)-2}-1}{a-1},$ and by Euler's theorem, $a^{\phi(m)}\equiv1\pmod m.$ Now, because $(a,m)=1$, we have $a^{\phi(m)-2}\equiv a^{-2}\pmod m,$ $a^{\phi(m)-2}-1\equiv a^{-2}-1\pmod m,$ and because $(a-1,m)=1$, $\frac{a^{\phi(m)-2}-1}{a-1}\equiv\frac{a^{-2}-1}{a-1}\pmod m,$ $1+a+a^2+\cdots+a^{\phi(m)-1}\equiv\frac{a^{-2}-1}{a-1}\pmod m.$ However, I get stuck here. Is there a way to show that the RHS of that last expression is congruent to zero modulus $m$? Thanks in advance!
Note: I really do not know if I am tackling this problem correctly to begin with.