A proof of a special case of Dirichlet's theorem.

Question

I came up with a proof of the special case of Dirichlet's theorem, stated below:

For any integer $n>0$, there are infinitely many integers $k>0$ such that $kn+1$ is a prime.

Proof. Assume $n>1$ and choose a prime divisor $p$ of $\Phi_n(n)$, where $\Phi_n$ is the $n^{th}$ cyclotomic polynomial, then $n^n\equiv 1$ modulo $p$. Let $o(n)$ be the order of $n$ in $(\Bbb Z/p\Bbb Z)^{\times}$, then $o(n)|n$. As $(\Bbb Z/p\Bbb Z)^{\times}$ is cyclic of order $p-1$, one has $p\equiv 1$ modulo $o(n)$,

If $o(n)<n$, then $p$ divides $n^{o(n)}-1=\prod_{d|o(n)}\Phi_d(n)$ thus $p$ must divide some $\Phi_d(n)$ with $d<n$ and $d|n$. In other words, $x^n-1 = \prod_{d|n}\Phi_d$ is not separable over $\Bbb F_p$. This is impossible by field theory unless $p|n$, which implies that $\Phi_n(n)\equiv \Phi_n(0) = 1$ modulo $p$, contradiction.

So $o(n)=n$ and we obtain one prime $p\equiv 1$ modulo $n$, say $p=kn+1$. Take sufficient large integer $m$ such that $n^m>k$ and there is a new prime $p'=k'n^{m+1}+1=(k'n^{m})n+1$ for some $k'$. This construction gives infinitely many primes of the form $kn+1$.$\square$

I believe this is a correct proof but it seems a little too elementary. Is it correct? Can it even be suitably generalized?