How many elements there are that generate $\mathbb{F}_{3^{60}}$ over $\mathbb{F}_{3^6}$? Let $F/k$ be an extension of fields with $|k|=3^6$ and $|F|=3^{60}$, how many elements $\alpha\in F$ are there with $k(\alpha)=F$?
 A: For any prime power $q$ there are only two intermediate fields between $K=F_q$ and $L=F_{q^{10}}$, namely $M_1=F_{q^2}$ and $M_2=F_{q^5}$. This follows either from the general theory of finite fields or from Galois theory. The Galois group $Gal(L/K)$ is cyclic of order 10, and hence its only non-trivial subgroups are cyclic groups of orders five and two respectively, and then $M_1$ and $M_2$ are the corresponding fixed fields.
The element $x\in L$ generates $L$ over $K$, unless $x\in M_1$ or $x\in M_2$. The intermediate fields $M_1$ and $M_2$ intersect at $K$, so this happens for
$$
N=|M_1\cup M_2|=|M_1|+|M_2|-|M_1\cap M_2|=q^5+q^2-q
$$
elements $x$.
Therefore the correct answer is that $L=K(x)$ for 
$$q^{10}-N=q^{10}-q^5-q^2+q=3^{60}-3^{30}-3^{12}+3^6$$
choices of $x$.
The general formula involves the Möbius function via the usual exclusion/inclusion business.
A: The elements of $\mathbb F_{q^s}$ which generate $\mathbb F_{q^s}$ are the elements that are not in any strict subfield of $\mathbb F_{q^s}$. These subfields are all the $\mathbb F_{q^r}$, with $r$ dividing $s$.
You can compute the cardinal of $$\bigcup_{r | s, r < s} \mathbb F_{q^r}$$
using inclusion-exclusion principle and the fact that $\mathbb F_{q^b} \cap \mathbb F_{q^a} = \mathbb F_{q^{\gcd(a, b)}}$.
For your particular example, we have $q = 3^6$ and $s = 10$. The strict divisors of $s$ are 1, 5, and 2. Thus the number you are looking for is
$$ q^{10} - (q^5 + q^2 + q^1) + 2q^1 = 42391158275215997623161807840$$
