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) ?

$\mathbb{Z}_p^*$ is a cyclic group of order $p-1$, so by Lagrange theorem if $a^{\frac{p-1}{2}} = y$ we have $$y^2 = (a^{\frac{p-1}{2}})^2 = a^{p-1} = 1$$ So y is a root of $x^2 -1 \in \mathbb{Z}_p[x]$, but $\mathbb{Z}_p$ is a field and so there are at most $2$ roots, i.e. $\lbrace 1, -1\rbrace$.
So  $y = a^{\frac{p-1}{2}} = \pm 1 $. 
Moreover if you define $$\phi : \mathbb{Z}_p^* \to \lbrace 1, -1 \rbrace$$ $$\gamma \to \gamma^{\frac{p-1}{2}}$$ this is a surjective ( because $p-1 > \frac{p-1}{2}$ ) homorphism group, and so the cardinality of the kernel, call it K,  is $\frac{|\mathbb{Z}_p^*|}{2} = \frac{p-1}{2}$.
The squares $\mod p \ $ in $\mathbb{Z}_p^*$ are $\frac{p-1}{2}$, because if $\alpha, \beta \in \mathbb{Z}_p^* $ and $\alpha^2 = \beta^2$ this implies $\alpha = \pm \beta$.
So we have that $$H = \lbrace \alpha^2 \mid \alpha \in  \mathbb{Z}_p^* \rbrace \subseteq K$$ and $|H| = |K|$. This means that $H = K$ and so for $a \in \mathbb{Z}_p^*$ $$  a^{\frac{p-1}{2}} = 1 \Leftrightarrow a \ \text{is a square}$$
