Given that $p$ is a prime and $n$ is a natural number, I want to know how many $\alpha$ are in $\mathbb F_{p^n}$ such that $\mathbb F_p (\alpha) = \mathbb F_{p^6}$.
I know that $[\mathbb F_{p^6} : \mathbb F_p] = 6$, so $\mathbb F_p (\alpha) = \mathbb F_{p^6}$ would happen if and only if the irreducible polynomial of $\alpha$ with coefficients in $\mathbb F_p$ has a degree of $6$, but not sure how to proceed from here.
To long for a comment, let me know if there are mistakes in my argument.
For $a,b\in \mathbb N$ it holds $\mathbb F_{p^a} \subset \mathbb F_{p^b} \Leftrightarrow a|b$, so if $n$ is not a multiple of $6$ the field $\mathbb F_{p^6}$ is not contained in $\mathbb F_{p^n}$.
If $n$ is a multiple of $6$, fix an algebraic closure $\bar{\mathbb F}$ of $\mathbb F_p$. The elements of $\mathbb F_{p^n}$ are exaclty the roots of $X^{p^n}-X\in \mathbb F[X]$ in $\bar{\mathbb F}$, and the elements of $\mathbb F_{p^6}$ are exaclty the roots of $X^{p^6}-X$. So the elements $\alpha\in \mathbb F_{p^n}$ generating for $\mathbb F_{p^6}$ are the elements in this field which are not contained in a proper subfield: $\mathbb F_{p}$, $ \mathbb F_{p^2}$ and $\mathbb F_{p^3}$, therefore there are $$p^6-(p^3-p)-(p^2-p)-p= p^6-p^3-p^2+p$$ generating elements, since $\mathbb F_{p^3}\cap \mathbb F_{p^2}= \mathbb F_{p}$.