Why is it true that $F_{q^n} = F_q(\alpha)$ where $\alpha$ is the primitive element of $F_{q^n}$?

Question

Since $\alpha$ is the primitive element of $F_{q^n}$ then $F_{q^n} = \{0, \alpha, \alpha^2,\cdots, \alpha^{q^{n -2}} , 1\}$. Then how $F_q(\alpha)$ is equivalent to $F_{q^n}$? Because what I understand $F_q(\alpha)$ is extension of $F_q$ by adjoining $\alpha$ alone.