The question asks to show that if p is an odd prime, and r is a primitive root of p, then ind_r (p-1) = (p-1)/2

I don't even know where to start -- r^(p-1) = r^(phi(p)) which is congruent to one modulo p. Writing this out, I would have that r^(p-1) = kp +1.

I'm just starting on modular arithmetic today, so forgive me for my lack of knowledge...