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.
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 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.