Show that if $G$ is a cyclic group of order $p^n-1$ ($p-$ prime) then every element of G can be written in the form $a^p$, $a \in G$
I know that if $a \in G$ then $$a^{p^n-1}=e$$ where $e $ is the unit element in $G$
$$a^{p^n}=a \Rightarrow a^{p^{n+1}}=a^p$$ Now since $(p^n-1,p^{n+1})=1$ then $a^p$ generates the group $G$
Is this correct?
Let $r,s$ be coprime positive integers. Then if $G$ is finite of order (or just exponent) $r$, every element of $G$ can be written as $a^s$ with $a\in G$.
To see this, not that there are integers $u,v$ such that $ur+vs=1$. Now given $g\in G$, we have $$ g=g^1=g^{ur+vs}=(g^r)^u(g^v)^s=(g^v)^s$$ as desired.