For the isomorphism of $U_8$ (where $U_8 = \{ z \in \mathbb{C} | z^8 = 1 \}$) with $\mathbb{Z}/8\mathbb Z$ in which $\zeta =e^{i2\pi/8} \mapsto 5$ and $\zeta \cdot \zeta = 5 +_8 5 =2$
Why is $\zeta^0 = 1$?
I cannot figure out how we get 0 in this.
EDIT: Directly quoting the problem
It can be shown that there is an isomprhism of $U_8$ with $\mathbb{Z}_8$ in which $\zeta = e^{i\pi/4} \leftrightarrow 5$ and $\zeta^2 = 2$. For $m=0$, we have $\zeta^0 = 1$ and for $m = 3$ we have $\zeta^2 \cdot \zeta =2 +_8 5=7$ and similarly $\zeta^4 = \zeta^2 \zeta^2 = 2 +_8 2 = 4$
ADDED QUESTION
Why is $2+_8 5= 7$? Why isn't it $2 + 5 - 8 = -1 $?