Prove that any finite cyclic group with more than two elements has at least two different generators.
Attempt: Let $G$ be a finite cyclic group with 3 elements, say $e,g,g^{-1}\in G$ such that $e=$identity element. If $G$ is a cyclic group then it must have a generator, say $g$. If $g$ generates $G$ then so does $g^{-1}$ since $o(g)=n$ implies $g^n=e$, and $g^{-n}=(g^{-1})^n=e.$ Therefore, $G$ has at least two generators; $g$ and $g^{-1}.$
(I know this is a possible duplicate but I am asking a question specifically about my proof. Is it correct?)
As mentioned in the comments $g^{-n} = e$ does not imply that $g^{-1}$ has order $n$, only that the order of $g^{-1}$ is a divisor of $n$. You should base your proof that the number of elements of order $n$ is $\Phi(n)$ (the Euler totient function).