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.
Suppose that the map $x\mapsto x^p$ is non-injective: $a^p=b^p, a\ne b$. Since $p, p^n -1$ are coprime there exist $u,v$ such that $up+v(p^n-1)=1$. Then $a=a^{up+v(p^n-1)}=b^{up+v(p^n-1)}=b$, a contradiction. So the map is injective. Then it is surjective. QED We are using the fact that in a group of order $k$ we have $x^k=1$ for every $x$.
Conceptually, by below: $\,k\!=\!p\,$ is coprime to $\color{#0a0}{f= p^n\!-\!1}\,$ so we can take $p$'th roots: $\,g = (g^{1/p})^p$.
Theorem $ $ [Compute $k$'th root by raising to power $\frac{1}k\!\pmod{\!f}\,$ if $\,k\,$ is coprime to $\color{#0a0}{{\rm period}\ f}$]
Suppose that $\ \color{#0a0}{a^f} = 1 = \color{#0a0}{b^f},\ $ and that $\ \color{#c00}{1/k}_f:= k^{-1}\equiv k'\pmod{\!f},\, $ so $\, \color{#90f}{kk'} = 1 \!+\! jf.\, $ Then
$$ \bbox[10px,border:1px solid #c00]{\large a^{\large\color{#c00} k}= b \iff a = b^{\large (\color{#c00}{1/k})_{f}} }\qquad\qquad$$
Proof $\ (\Rightarrow)\ \ \ b = a^{\large k}\,\Rightarrow\, b^{\large k'}\! = a^{\large\color{#90f}{kk'}}\! = a^{\large 1+fj} = a(\color{#0a0}{a^{\large f}})^{\large j} = a.\ $ $(\Leftarrow)\ $ Same way $\ (a = b^{\large k'})^{\large k}$
Remark $ $ The proof was excerpted from this answer in an elementary number theory context (which might prove helpful as motivation). If you have good knowledge of (cyclic) groups then you may find it instructive to polish it in that language.
