I am trying to find the primitive roots of $ p^{n}$ and know that $\phi(p^{n})=(p-1)p^{n-1} $. However, I don't know how to find the prime divisors of $ \phi(p^{n}) $ ?
I am considering finding the primitive roots of p but not sure if I would be able to generalise to the primitive root of $p^{n}$. If I can then in this case $\phi(p)=p-1 $, but I am still not too sure what the prime divisors of this are ?
We want to find the prime divisors of $(p-1)p^{n-1}$ for some prime $p$. The answer is, there exist at least two prime divisors, namely $2$ and $p$, for all primes $p$ with $n > 1$.
Proof: Of course, $p\mid (p-1)p^{n-1}$ because $(*)\,\,p\mid p^{n-1}$. And therefore, since $\forall p > 2$, we have that $2\mid p - 1$ because every prime $p > 2$ is odd, then also $2\mid (p-1)p^{n-1}$, even if $p = 2$ because of statement $(*)$. $$\therefore \{2, p\}\mid (p-1)p^{n-1}.$$ What if $n = 1$? Then $p\mid p^{1-1} = p^0 = 1$, which is a contradiction. So, as a consequence, $$\{2, p\}\mid (p-1)p^{n-1}\tag{$n>1$}.$$ And since $p\geqslant 2$ with $p_1 = 2$ (this is to say, $2$ is the first prime), it follows that $$\{2, p\}\mid (p-1)p^{n-1}.\tag*{$\begin{align} (\forall p\text{ prime}) \\ (n > 1)\end{align}$}$$ We thus obtained as desired. $\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\,\,\Box$
Nota Bene: By this proof, we can be sure that $2$ and $p$ divide $(p-1)p^{n-1}$ but this does not mean that $2$ and $p$ are the only divisors. It just means that $(p-1)p^{n-1}$ will always have at least these divisors.