Compute $d(n) = [\mathbb{F}_3[a] : \mathbb{F}_3]$ where $a\in K^\times$ has order $n$ and $K$ is a field with $81$ elements.

Question

I know that the group of units $K^\times$ is cyclic of order $80$, so the possible values of $n$, the order of $a$, are $1,2,4,5,8,10,16,20,40,80$, the divisors of $80$. Now I am tempted to think that $d(n)=n$ since if $a$ has order $n$, then the $\mathbb{F}_3$-vector space $\mathbb{F}_3[a]$ has basis $1,a,a^2,...,a^{n-1}$. Is this correct? I feel like I am missing something.

Notation: $[K:F]$ is the dimension of $K$ as an $F$-vector space