Definition A number field $K$ is a finite degree extension of $\mathbb Q$. This means that every number field is generated in by rational combinations of a finite basis $\bar e$,
Theorem Every element of $K$ is algebraic.
Proof: Take an element $\alpha$ and consider expressing each of $1,\alpha,\alpha^2,\cdots,\alpha^d$ rationally in terms of the basis of the number field. These can be thought of as $d+1$ vectors in a dimension $d$space, hence a nontrivial linear relation between them exists - this gives rise to a polynomial for which $\alpha$ is a root.
Theorem Every number field is generated by a single element. This implies $K = \mathbb Q[X]/(p(X))$ for some polynomial $p(X)$.
Proof: Since every element is expressible in terms of the basis we have $K = \mathbb Q(\bar e)$. By induction on the length of the generating set $\bar e$ the theorem follows from the lemma:
Lemma For algebraic numbers $\alpha,\beta$ there exists an algebraic $\gamma$ such that $\mathbb Q(\alpha,\beta) = \mathbb Q(\gamma)$.
Before proving this here is a result due to Artin:
Theorem (Primitive element theorem) If there are finitely many fields between $K$ and $\mathbb Q$ then, $K$ is simple.
Proof: Consider the fields $K_c = \mathbb Q(\alpha + c \beta)$, there are infinitely many but by the hypothesis there must be $c \not = c'$ with $K_c = K_{c'}$. In this field it is an easy exercise to recover $\alpha$ and $\beta$.
There really are finitely many fields below a number field $K$, but I don't see any simple proof of this. The lemma can be proved by a much more clever technique inspired by the previous argument (Artin - Algebra):
Let's take the minimal polynomials $p(\alpha) = 0$, $q(\beta) = 0$ and construct $\gamma_c = \alpha + c \beta$ for rational $c$. Define an auxilliary polynomial $h_c(x) = p(\gamma_c - c x)$ so that $h_c(\beta) = 0$. Now consider the gcd $g = (q,h_c)$.
Clearly $g(\beta) = 0$, but it may have other roots. If we can pick $c$ so that it does not have any other roots it will be a linear polynomial: $g = (x - \beta)$, this would prove $\beta \in \mathbb Q(\gamma_c)$.
$g(x) = 0$ iff simultaneously $q(x) = 0$ and $h_c(x) = 0$. i.e. $x = \beta_i$ and $\gamma_c - c x = \alpha_i$. Given those, $c = - \frac{\alpha - \alpha_i}{\beta - \beta_i}$. So simply pick $c$ avoiding these finitely many 'bad' values.
Warning! Seperability of $\beta$ was required for the $c$ expression to be well formed.
Why would there be only finitely many fields between $K$ and $\mathbb Q$? If this could be proved on it's own then we could use the primitive element theorem rather than that very difficult lemma.
There is an approach using seperability based on the following:
Theorem Every finite seperable extension is simple.
This seems to be the hardest and most technical part. There is a proof in Algebra - Chapter 0. This again involves the linear form $\alpha + c \beta$ - but it also makes use of algebraic closures and a tricky argument with polynomials and automorphisms.
Seperability is important and not too difficult to show: If $f$ has repeated roots it can be reduced since $\gcd(f,f')$ divides it. There is also this:
Lemma If $f$ is seperable then $f$ and $f'$ are relatively prime.
The theorems also hold in the finite field case but are an easy consequence of the group of units of a finite field being cyclic.