Is there an "elementary" proof of the infinitude of completely split primes?

By the primitive element theorem, $K=\mathbb{Q}(\alpha)$ for some nonzero $\alpha \in K$, and we may assume that the minimal polynomial $f(x)$ of $\alpha$ has integer coefficients. Let $\Delta$ be the discriminant of $f$. Since $K/\mathbb{Q}$ is Galois, a prime $p \nmid \Delta$ splits completely in $K$ if and only if there is a degree $1$ prime above $p$, which is if and only if $p | f(n)$ for some $n \in \mathbb{Z}$. Suppose that the set $P$ of such primes is finite. Enlarge $P$ to include the primes dividing $\Delta$. Let $t$ be a positive integer such that $\operatorname{ord}_p t> \operatorname{ord}_p f(0)$ for all $p \in P$. For any integer $m$, we have $f(mt) \equiv f(0) \;(\bmod \; t)$, so $\operatorname{ord}_p f(mt) = \operatorname{ord}_p f(0)$ for all $p \in P$. But $f(mt) \to \infty$ as $m \to \infty$, so eventually it must have a prime factor outside $P$, contradicting the definition of $P$.

I don't think that Lenstra's and Stevenhagen's article

Primes of degree one and algebraic cases of Čebotarev's theorem, L'Enseign. Math. 37 (1991), 17-30

has been mentioned yet. It is available online here.

There's an old easy proof of the fact that there are infinitely many primes $p$, $p \equiv 1 \bmod n$: Let $\Phi_n(X)$ be the $n$-th cyclotomic polynomial. Show that $\Phi_n(X)$ has a root in $\mathbb{F}_p$ if and only if $p \equiv 1 \bmod n$. Do as in Euclid's proof of the infinitude of primes: if $p_1, \dots, p_r $ are primes $\equiv 1 \bmod n$ then consider $\Phi_n(n p_1 \dots p_r)$. It's bigger than 1 and not divisible by any of the $p_i$ or any prime dividing $n$. So this shows the statement for cyclotomic fields.