Prove Proposition (6.18).
Let's remind ourselves of Proposition (6.18).
Let $r$ be a primitive root modulo $p$ where $p$ is prime. Then $r^m \pmod p$ is also a primitive root modulo $p$, provided $\gcd (m, p-1) = 1$.
Since $r$ is a primitive root modulo $p$, then we know the least positive integer $k$ such that $r^k \equiv 1 \pmod p$ is $k=\phi(n)=p-1$.
Let's now consider the least positive integer $k$ such that $(r^m)^k \equiv 1 \pmod p$.
$$ (r^m)^k \equiv r^{mk} \equiv 1 \pmod p $$
Since $p-1$ is the least positive index of $r$ to give a result congruent to 1, this means
$$ p - 1 \mid mk $$
If $\gcd(m,p-1)=1$ then $p-1 \not \mid m$ and so $p-1 \mid k$. The least positive $k$ for which this is true is $k=p-1$.
This means the least positive integer $k$ such that $(r^m)^k \equiv 1 \pmod p$ is $k=p-1\phi(p)$, and so $r^m \pmod p$ is also a primitive root modulo $p$.
Note: the author's solution uses Corollary (6.9) which states that if $a$ modulo $n$ has order $k$, then $a^s$ has order $k$ if and only if $\gcd(s,k) =1$. Given the order of $r \pmod p$ is $p-1$, and since $\gcd(m, p-1)=1$, the corollary tells us $r^m$ has order $p-1$.