Consider the residues modulo $r$ of $p, p^2, p^3, \dots, p^{r-1}$.
Claim: $p^m \equiv 1 \bmod r$ for some $m, 1 \leq m \leq r-1$.
The proof is by contradiction. All $r-1$ residues belong to the set $\{1,2,\ldots, r-1\}$, and so if none of them equals $1$, then two residues must have the same value and thus $r$ is a divisor of $p^i - p^j = p^j(p^{i-j}-1)$. Since $\gcd(r, p^j) = 1$, $r$ must be a divisor of $(p^{i-j}-1)$. that is, $p^{i-j} \equiv 1 \bmod r$ in contradiction of the assumption that none of the residues equals $1$. $\quad\qquad\Box$
Let $m$ denote the smallest positive integer such that $r$ divides $p^m-1$. There exist irreducible polynomials of degree $m$ in $\mathbb F_p[x]$ (for the exact number, see, for example, here), and if $g(x)$ is such a polynomial, then $E = \mathbb F_p[x]/g(x)$ is an extension field of $\mathbb F_p$. $|E| =p^m$ and its nonzero elements constitute a cyclic group of order $p^m-1$ under multiplication. This group has elements of all orders $n$ that divide $p^m-1$. Since $r$ is a divisor of $p^m-1$, $E$ contains a primitive $r$-th root of unity, which is what you have to show. $\quad\qquad\Box$