We can prove something more generic.
Using this, if ord$\displaystyle _{p^s}a=d,$ ord$\displaystyle _{p^{s+1}}a=d$ or $p\cdot d$ where $p$ is odd prime and ord$_na$ is the multiplicative order of $a\pmod n$
By definition: $\ a$ is a primitive root $\iff a\,$ has order $\,p\!-\!1.\ $ Now apply the fundamental
Order Test $\ \,a\,$ has order $\,n\iff a^n \equiv 1\,$ but $\,\color{#0a0}{a^{n/q} \not\equiv 1}\,$ for every prime $\,q\mid n$
Proof $\,\ (\Leftarrow)\,\ $ Let $\,a\,$ have $\,\color{#c00}{{\rm order}\ k}.\,$ Then $\,k\mid n\,$ (proof). $ $ If $\:k < n\,$ then $\,k\,$ is proper divisor of $\,n\,$ therefore $\,k\,$ arises by deleting at least one prime $\,q\,$ from the prime factorization of $\,n,\,$ hence $\,k\mid n/q,\,$ say $\, kj = n/q,\ $ so $\ \color{#0a0}{a^{\large n/q}} \equiv (\color{#c00}{a^{\large k}})^{\large j}\equiv \color{#c00}1^{\large j}\equiv\color{#0a0}1\,$ contra $\rm\color{#0a0}{hypothesis}$. So $\,k=n.$ $\ (\Rightarrow)\ $ Clear.
Best Answer
Note that $\left(r^{(p-1)/2}\right)^2\equiv 1\pmod{p}$, so $r^{(p-1)/2}$ is a solution of the congruence $x^2\equiv 1\pmod{p}$.
The congruence $x^2\equiv 1\pmod{p}$ has precisely two solutions, $x\equiv 1\pmod{p}$ and $x\equiv -1\pmod{p}$.
Since $r^{(p-1)/2}\not\equiv 1\pmod{p}$, we must have $r^{(p-1)/2}\equiv -1\pmod p$, so $-1$ has index $(p-1)/2$.
Another way: If you want to just use "algebraic" properties of indices, let $d$ be the index of $-1$. Then the index of $(-1)^2$ is congruent to $2d$ modulo $p-1$. (Index of a product is, sort of, the sum of the indices.) But the index of $(-1)^2$ is $p-1$. so $2d\equiv 0\pmod{p-1}$ and $1\le d\lt p-1$. This forces $d=\frac{p-1}{2}$.