Why does it follows that $\alpha \in (\mathbb Z_p^*)^2 \iff \alpha^{(p-1)/2} = 1$ from proved results? Suppose I
've proved the following, where $(\mathbb Z_p^*)^2$ denotes the set of unit residue classes modulo $p$.

Why does it then follows that $\alpha \in (\mathbb Z_p^*)^2 \iff \alpha^{(p-1)/2} = 1$ and $\alpha \notin (\mathbb Z_p^*)^2 \iff \alpha^{(p-1)/2} = -1$ ?
Should I look at the contrapositive of each statement $\alpha \in (\mathbb Z_p^*)^2 \Leftarrow \alpha^{(p-1)/2} = 1$, $\alpha \notin (\mathbb Z_p^*)^2 \Leftarrow \alpha^{(p-1)/2} = -1$, or is there a more "direct" way also using (i) ?

The direction $(\Rightarrow)$ is $\rm(ii),(iii)$ in the theorem. The opposite direction follows because the set of nonzero squares and nonsquares form a partition of $\,\Bbb Z_p^*.\,$ Thus if $\,\alpha^{(p-1)/2} = 1\,$ then $\,\alpha\,$ is either a square or a nonsquare, but it cannot be a nonsquare since those map to $-1,\,$ so it must be a square.   
Thus it boils down to the set-theoretic fact that a map on a set induces a partition of the set into the map's fibers (a.k.a. level-sets or preimages). See also this question.
