In the book we can read
Lemma 5.14. Let $K(\alpha) : K$ be a simple algebraic extension, let the minimal polynomial of a over $K$ be $m$, and let $\partial m = n$. Then $ \{ 1 , \alpha ,\dots \alpha^{n-1} \}$ is a basis for $K(\alpha)$ over $K$.
Proof The theorem is a restatement of Lemma 5.9.
Of course I reread such lemma, that states,
Lemma 5.9. Every polynomial $\alpha\in K[t]$ is congruent modulo $m$ to a unique polynomial of degree $< \partial m$.
This lemma is not even about dimension. I think there is some error, if not, I am unable to restate lemma 5.9 to arrive to lemma 5.14. In any case I'd like to understand some proof. I can figure out in the examples that a basis for the extension is that the lemma states (we can get $\alpha^2$, $\alpha^3$ etc from the product, but the e.v. operations only allow sums), but a more formal proof will be welcome.
I'm not sure either what they meant by a restatement of lemma $5.9$. Let's look at lemma $5.14$. Consider the minimal polynomial of $\alpha$ over $K$, denoted by $f^{\alpha}_K := f = \sum^n_{i=0} b_i x^i \in K[x]$, where $b_n =1$.
The intuitive idea behind lemma $5.14$ is that you can express any power of $\alpha$ larger than (or equal to) $n$ by the elements $\{1,\alpha, \alpha^2,...\alpha^{n-1}\}$, because of the relation given by $f$ (you can rewrite the expression $f(\alpha) = 0$ to $\alpha^n = \sum^{n-1}_{i=0}b_i\alpha^i$). Therefore they already generate $K(\alpha)$ over $K$. The linear independence for the elements $\{1,\alpha, \alpha^2,...\alpha^{n-1}\}$ follows from the fact that the degree of the extension $K \subset K(\alpha)$ is exactly equal to the degree of $f$ (which is equal to $n$).