Let $F$ be a field having $5^n$ elements .Also $F$ has an element which satisfies $x^{5^n}=1$ such that $x\neq 1$. Find $n$ .
My try:
Let $x\in F $ satisfy $x^{5^n}=1$ .Obviously the group $F^{*}=F\setminus \{0\}$ is cyclic. Consider the cyclic group generated by $x$ i.e $\langle x\rangle $. Also $|F^{*}|=5^n-1$
Let $o(x) =t$ .Now $t$ divides $5^n-1$ since order of an element divides order of the group. Since $x^{5^n}=1\implies t $ divides $5^n$.
Thus $t$ divides both $5^n$ and $5^n-1\implies x=1$ but $x\neq 1$ .Hence no such $n$ exists.
Is the solution correct ?Am I missing something.Please help