In the following corollary in Artin,
For every positive integer $r$, there exists an irreducible polynomial of degree $r$ over the prime field $\mathbb{F}_p$.
Proof: According to (d),there is a field $K$ of order $q = p^r$. Its degree $[K:F]$ over $F = \mathbb{F}_p$ is $r$. According to (c), the multiplicative group $K^×$ is cyclic. It is obvious that a generator $\alpha$ for this cyclic group will generate $K$ as extension field, i.e., that $K = F(\alpha)$. Since $[K: F] = r$, $\alpha$ has degree $r$ over $F$. So $\alpha$ is the root of an irreducible polynomial of degree $r$.
I don't understand the following sentence
It is obvious that a generator $\alpha$ for this cyclic group will generate $K$ as extension field, i.e., that $K = F(\alpha)$.
Why is it true?
It is obvious that $F(\alpha)\subseteq K$ since $\alpha\in K$. On the other hand, if $x\in K$, either $x=0$ and it is obvious that $x\in F(\alpha)$, either $x\in K^*$ so that $x=\alpha^n$ for some $n\in\mathbb{Z}$ which means that $x\in F(\alpha)$ i.e. $K\subseteq F(\alpha)$.