I am not sure how to prove this:
If a has order t modulo a prime p, show that a^(t-1)+a^(t-2)+...+a+1 == 0(mod p).
$\displaystyle 1+a+\cdot\cdot\cdot+a^{t-2}+a^{t-1} = \frac{1-a^t}{1-a} \equiv 0\mod{p} $ since $\displaystyle a^t\equiv 1 \mod{p} $.
Note that it's necessary that $\displaystyle \text{ord}_p(a)>1 $, otherwise $\displaystyle (1-a)^{-1} $ doesn't exist modulo $\displaystyle p $.