The special case of Dirichlet's theorem for primes in arithmetic progressions for primes congruent to $1$ modulo $q$ where $q$ itself is a prime can be established by the following elegant modification of Euclid's proof.

Let . Let $n>1$ be an integer, and suppose $p\mid f(n)$ . Then $n^q\equiv 1\pmod p$ which implies by Lagrange's theorem that either $q\mid p-1$ or $n\equiv 1\pmod p$ . In other words, every prime divisor of $f(n)$ is congruent to $1$ modulo $q$ unless $n$ is congruent to $1$ modulo that divisor.

Suppose there are only finitely many primes that are congruent to $1$ modulo $q$ . Let $P$ be twice their product. Note that $P\equiv 2\pmod q$ . Let $p$ be any prime divisor of $f(P)$ . If $p\equiv 1\pmod q$ , then $p\mid P$ which contradicts $f(P)\equiv 1\pmod P$ . Therefore, by the above $P\equiv 1\pmod p$ . Therefore . Since $q$ is prime, it follows that $p=q$ . Then $P\equiv 1\pmod p$ implies $P\equiv 1\pmod q$ . However, that is inconsistent with our deduction that $P\equiv 2\pmod q$ above. Therefore the original assumption that there are only finitely many primes congruent to $1$ modulo $q$ is false.

References

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Henryk Iwaniec and Emmanuel Kowalski.
Analytic Number Theory, volume 53 of AMS Colloquium Publications.
AMS, 2004.