Prove Proposition (6.17).
Let's remind ourselves of Proposition (6.17).
Let $n$ have a primitive root and $a$ and $n$ be relatively prime. The congruence $$x^m \equiv a \pmod n)$$ has a solution if and only if $a^{\phi(n)/g} ≡ 1 \pmod n$ where $g= \gcd(m, \phi (n))$. Additionally, there are exactly $g$ incongruent solutions.
Applying Proposition (6.15) and (6.16) to $x^m \equiv a \pmod n)$ gives us
$$ m \; \text{ind}_r(x) \equiv \text{ind}_r(a) \pmod {\phi(n)} $$
where $r$ is the primitive root of $n$.
This is now a linear congruence, and by Proposition (3.16), has a solution for $\text{ind}_r(x)$ if and only if $g = \gcd(\phi(n),m)$ divides $\text{ind}_r(a)$, and if so, there are $g$ incongruent solutions.
Let's examine the condition $g = \gcd(\phi(n),m)$ divides $\text{ind}_r(a)$. This means, for some integer $k$, we have $ \text{ind}_r(a) = k g $
That is
$$ r^{kg} \equiv a \pmod n $$
Since $g$ is a divisor of $\phi(n)$, then $/phi(n)/g$ is an integer. We can raise both sides of the congruence to this integer.
$$ (r^{kg})^{\phi(m)/g} \equiv a^{\phi(n)/g} \pmod n $$
Simplifying, and using that $r^{\phi(n)}\equiv 1 \pmod n$.
$$ (r^{\phi(n)})^k \equiv 1 \equiv a^{\phi(n)/g} \pmod n $$
This is the desired conclusion.