October 26th
Today I learned an elementary proof that the set of primes which split completely in a Galois extension $K/\QQ$ of degree $n$ have density $1/n,$ only assuming some minor analytic facts about the Dedekind zeta function. Namely, this weaker statement does not require use of the Chebotarev Density Theorem. Basically, recall\[\zeta_K(s)=\sum_{I\subseteq\mathcal O_K}\frac1{\op N(I)^s}=\prod_\mf p\frac1{1-\op N(\mf p)^{-s}}.\]Noting that $\zeta_K(s)$ has a simple pole of rank $1$ at $s=1,$ we may say that $(s-1)\zeta_K(s)$ approaches some positive real number as $s\to1^+.$ We use this to get the result. In particular, we get to write\[\lim_{s\to1^+}\frac{\log(\zeta_K(s))}{\log\left(\frac1{s-1}\right)}=1+\lim_{s\to1^+}\frac{\log((s-1)\zeta_K(s))}{\log\left(\frac1{s-1}\right)}=1.\]Because this holds for $K=K$ and for $K=\QQ,$ we can divide the limits corresponding to each to get\[\lim_{s\to1^+}\frac{\log(\zeta_K(s))}{\log(\zeta_\QQ(s))}=1.\]But, akin to Dirichlet's Theorem, we can also expand\[\log(\zeta_K(s))=\log\left(\prod_{\mf p}\frac1{1-\op N(\mf p)^{-s}}\right)=\sum_{\mf p}\sum_{k=1}^\infty\frac1{k\op N(\mf p)^{sk}}\]because $-\log(1-x)=\sum_kx^k/k.$ We're going to get our result because the only terms that we care about in the above sum come from the rational primes that split completely.
This is an absolutely convergent series (all the terms are positive), so we are a bit liberal with our rearranging. In particular, we order this by rational primes $p$ so that this is\[\log(\zeta_K(s))=\sum_p\sum_{\mf p/(p)}\sum_{k=1}^\infty\frac1{k\op N(\mf p)^{sk}}.\]Dealing safely with the outer sum to start, we note that $e(\mf p/p)$ and $f(\mf p/p)$ are constant. The number of primes is therefore $\frac n{e(\mf p/p)f(\mf p/p)},$ giving $\op N(\mf p)=p^{f(\mf p/p)}.$ It follows that\[\log(\zeta_K(s))=n\sum_p\frac1{e(\mf p/p)f(\mf p/p)}\sum_{k=1}^\infty\frac1{kp^{kf(\mf p/p)s}}.\]Now we reduce this sum. We remark that only finitely many primes $p$ have $e(\mf p/p) \gt 1,$ which will contribute a $-\log(1-p^{-f(\mf p/p)s})$ of inner sum, totalling only to $O(1)$ over all such (finitely many) primes. It follows\[\log(\zeta_K(s))=O(1)+n\sum_{e(\mf p/p)=1}\frac1{f(\mf p/p)}\sum_{k=1}^\infty\frac1{kp^{kf(\mf p/p)s}}.\]Further, note that we can also group all the $k \gt 1$ terms into\[\sum_{e(\mf p/p)=1}\frac1{f(\mf p/p)}\sum_{k=2}^\infty\frac1{kp^{kf(\mf p/p)s}} \lt \sum_{n=2}^\infty\sum_{k=2}^\infty\frac1{n^k}=\sum_{n=2}^\infty\frac1{n(n-1)} \lt \sum_{n=1}^\infty\frac1{n^2} \lt \infty.\]So it follows\[\log(\zeta_K(s))=O(1)+n\sum_{e(\mf p/p)=1}\frac1{f(\mf p/p)p^{f(\mf p/p)s}}.\]But again, we can group all $p$ with $f(\mf p/p) \gt 1$ by\[\sum_{\substack{e(\mf p/p)=1\\f(\mf p/p)\ge2}}\frac1{f(\mf p/p)p^{f(\mf p/p)s}} \lt \sum_{n=2}^\infty\frac1{n^2} \lt \infty.\]So we finally get to reduce our sum to\[\log(\zeta_K(s))=O(1)+n\sum_{\substack{e(\mf p/p)=1\\f(\mf p/p)=1}}\frac1{p^s}.\]However, this is a sum over $p$ completely splitting, which is equivalent to $e(\mf p/p)=f(\mf p)=1.$
To finish, we plug this into our limit from before, which tells us\[1=\lim_{s\to1^+}\frac{\log(\zeta_K(s))}{\log(\zeta_\QQ(s))}=\lim_{s\to1^+}n\cdot\dfrac{\displaystyle\sum_{p\text{ splits completely}}\frac1{p^s}}{\displaystyle\sum_p\frac1{p^s}},\]which rearranges to what we wanted, in terms of Dirichlet density.