'Simple' Proof: Infinitely Many Galois Fields of Fixed Degree TLDR; Is there an 'elementary' argument to prove the following:

Claim: Given an integer $d>1$, are there infinitely many distinct Galois extensions $K/\mathbb{Q}$ with $[K \colon \mathbb{Q}]= d$?

Elementary in the sense that students new to Field/Galois Theory could easily follow, and that can be explained 'quickly'?

'Brief Background': When discussing field extensions with students, I pointed out that every quadratic extension of $\mathbb{Q}$ is Galois but that this was certainly not generally true. A student then asked the next obvious question,

"What about cubic extensions? Are 'most' of them Galois or not? What about other fields? Are there only finitely many in some degrees? Are
fields in some degrees 'more' or 'less' Galois?"

Obviously, pointing out that 'most' extensions are not Galois, I stated that we can at least guarantee one for each degree - for instance, using the construction from the classic proof that every finite abelian group is a Galois group - and commented that there are infinitely many in each degree. Of course, counting them (up to fixed discriminant for example) is a difficult open problem.
I did not prove the latter statement for them, though in the moment I mentally prepped for the question which never came, probably due to them being 'side discussion overwhelmed.' But I think the question is a good one:

Claim: Given an integer $d>1$, are there infinitely many distinct Galois extensions $K/\mathbb{Q}$ with $[K \colon \mathbb{Q}]= d$?

Or more generally for fields $K_0$ with char $K_0=0$, are there infinitely many distinct Galois extensions $K/K_0$ with fixed degree, distinctness here meaning $K \cap K'= K_0$, where $K,K' \subseteq \overline{K_0}$ for some algebraic closure of $K_0$.
The Question: Is there a simpler argument using only 'basic' results from Field/Galois Theory that can be quickly presented and understood by students in their first pass at Abstract Algebra, if only for the case that $K= \mathbb{Q}$?
 A: I think this is not at all an elementary result; it really tells you something nontrivial about the absolute Galois group of $\mathbb{Q}$ (note that it's equivalent to the existence of infinitely many distinct open normal subgroups of index $d$). For example, I think it doesn't hold for $\mathbb{Q}_p$ because the absolute Galois group of $\mathbb{Q}_p$ is topologically finitely generated, which implies there are only finitely many Galois extensions with a fixed Galois group and hence finitely many Galois extensions with a fixed degree.
I think the fastest way to do this "from scratch" is to exhibit subextensions of cyclotomic extensions:

*

*For every prime $p$, prove that $\mathbb{Q}(\zeta_p)$ is Galois with Galois group $(\mathbb{Z}/p)^{\times}$.

*For every positive integer $n$, prove that there are infinitely many primes $p \equiv 1 \bmod n$ (this can be done in an elementary way using the cyclotomic polynomial $\Phi_n(x)$, we don't need Dirichlet's theorem), and hence infinitely many primes $p$ such that the Galois group of $\mathbb{Q}(\zeta_p)$ has a subgroup of index $n$, whose fixed field is an abelian Galois extension of $\mathbb{Q}$ of degree $n$, with Galois group $C_n$.

*Prove that for distinct primes these extensions are distinct (e.g. by computing their discriminants).

The existence of the cyclotomic extensions $\mathbb{Q}(\zeta_n)$ implies that the absolute Galois group surjects onto $\widehat{\mathbb{Z}}^{\times}$, the group of units of the profinite integers, which is an enormous group and in particular not topologically finitely generated.  (When we consider subextensions of $\mathbb{Q}(\zeta_p)$ we are considering the projection of this group to $(\mathbb{Z}_p / p \mathbb{Z}_p)^{\times}$.)

We can ask for elementary methods in each small degree at least. For degree $3$ we want to exhibit infinitely many cubics with Galois group $C_3$ with nonisomorphic splitting fields. This can be done without cyclotomic extensions by proving that an irreducible cubic has Galois group $C_3$ iff its discriminant is a square and exhibiting a family of irreducible cubics with square discriminants. This is already not entirely elementary. The discriminant of $x^3 - px + q$ is $\Delta = 4p^3 - 27 q^2$ so requiring that this is a square involves solving a Diophantine equation
$$4p^3 - 27q^2 = r^2.$$
Equivalently we want to exhibit infinitely many $p$ such that $4p^3$ can be represented by the quadratic form $r^2 + 27q^2$. Recognizing this as the norm of $r + 3q \sqrt{-3} \in \mathbb{Q}(\sqrt{-3})$, we can do this by taking $p = x^2 + 3y^2 = N(x + y \sqrt{-3})$ and writing $4 = N(1 + \sqrt{-3})$, which gives
$$4p^3 = N \left( (1 + \sqrt{-3})(x + y \sqrt{-3})^3 \right) = (x^3 - 9x^2 y - 9 xy^2 + 9y^3)^2 + 3 (x^3 + 3 x^2 y - 9 xy^2 - 3y^3)^2$$
which gives us our infinite family of solutions. Now we have to prove that infinitely many of these cubics are irreducible and that infinitely many of their splitting fields are distinct which seems a bit painful; at this point this seems to me like more work than using cyclotomic fields.
For degree $4$ things are easier because we can piggyback off of the degree $2$ case: we can consider the biquadratic extensions $\mathbb{Q}(\sqrt{a}, \sqrt{b})$ where $\gcd(a, b) = 1$.
For degree $5$ we want to exhibit infinitely many quintics with Galois group $C_5$ with nonisomorphic splitting fields, which without cyclotomic extensions seems hard again. In general things will be hard in prime degree $p$ since the Galois group must be $C_p$ and we have no intermediate fields to piggyback off of. For large $p$, and by "large" here I mean $p \ge 5$, I don't think it's clear a priori that there exists even one such extension.
