Find the sum of the $m$-th powers of the primitive roots mod p for a given prime p and a positive integer $ m$.

Question

Wikipedia has the result that Gauss proved that for a prime number p the sum of its primitive roots is congruent to $\mu(p−1)\pmod p$ in Article 81.

also see:Prove sum of primitive roots congruent to $\mu(p-1) \pmod{p}$

Now I have read other problem Titu Andreescu (Problems from the book) there a such hard problem for me:

Find the sum of the $m$-th powers of the primitive roots mod p for a given prime p and a positive integer $ m$.

when $m=1$ this is a Gauss result,so How to find for any $m-$ th power?Thanks

I tried If $g$ is a primitive root of $p$ $S$, is given by

$$\begin{align} S&=\!\!\!\!\!\sum_{\substack{k=1\\(k,p-1)=1}}^{p-1}\!\!\!\!\!\!g^{mk}\\&=\sum_{k=1}^{p-1}g^{mk}\sum_{\substack{d|k\\d|p-1}}\mu(d)\\&=\cdots\text{then I can't} \end{align} $$