A special case of Cheboratev's density theorem states that, for $K/\mathbb{Q}$ a Galois number field of degree $n$, then the rational primes that split completely in $K$ have density $1/n$.


Is there an elementary proof of this fact?


Actually I'm asking this only to apply it to $K=\mathbb{Q}(\zeta\_n)$ and get that the density is nonzero, and there is an elementary proof of this. However I'm also interested in the more general result.



Elementary proof of a special case of Chebotarev's density theorem
There does exist an elementary proof, and is given in many books on algebraic number theory. I think it is in Lang's book. I reproduce the proof below.


Consider $ \zeta \_K(s)=\prod \_{\mathfrak p}(1-\frac{1}{(N\mathfrak p)^{s}})^{-1}$ where $\mathfrak p$ runs over all prime ideals in the ring of integers in $K$. Therefore, 


$$\log (\zeta \_K(s))=\sum \_{\mathfrak p}\sum \_ {m\geq 1} \frac{1}{m(N\mathfrak p)^{ms}}$$


Since $\zeta \_K(s)=\frac{1}{s-1}(a\_0+a\_1(s-1)+\cdots)$ with $a\_0\neq 0$, it follows that $\frac{\log (\zeta \_K(s))}{\log \frac{1}{s-1}}$ tends to $1$ as $s$ tends to $1$ from the right. On the other hand, in this limit, only the term $m=1$ and $\mathfrak p$ of degree one over $\mathbb Q$ need be considered. Hence we get 
$$1= \lim \_{s \rightarrow 1} \frac{\sum \_{\mathfrak p} \frac{1}{(N\mathfrak p) ^s}}{\log \frac{1}{s-1}}.$$ Now, degree $1$ primes $\mathfrak p$ lie over rational primes $p$ which split completely in $K$, and over each $p$, there are $n=\deg(K/{\mathbb Q})$ primes $\mathfrak p$ of degree $1$. Hence we get 


$$1=\lim \_{s\rightarrow 1} n\left( \frac{\sum \_p \frac{1}{p^s}}{\log \frac{1}{s-1}}\right)$$ where the sum is over primes $p$ which split completely. This gives what you want.


