Applying Euler's Theorem to Prove a Simple Congruence 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.

Hint: From $a^{\phi(m)}-1$ is congruent to 0 mod m and is congruent to the $(a^{\phi(m)}-1)/(a-1)$ mod m
