Suppose there were a finite number of primes. We may label them $p_1, p_2, \ldots, p_n$. Let \[N=p_1 p_2 \ldots p_n + 1.\] Now for every prime $p_i,\,1\leq i\leq n$ that we just labelled, the remainder of $N$ divided by $p_i$ is 1. However, since $N$ is not prime, $N$ must be able to be factorized into a product of primes. What primes are in that factorization if it doesn't contain any of the primes we just listed out?

Hence, $N$ must have some new prime in its factorization or be a prime itself. This contradicts our assumption that there are a finite number of primes, and therefore an infinite number of primes must exist. $\blacksquare$

Suppose there are a finite number of primes that is of the form 4k-1 for some $k\in\mathbb{N}$. Again, we may label them $p_1, p_2, \ldots, p_n$. Let \[N=4 p_1 p_2 \ldots p_n - 1.\] Again, for every prime $p_i,\,1\leq i\leq n$ that we just labelled, the remainder of $N$ divided by $p_i$ is 1. Note that $N$ is not prime since it is of the form 4k-1 but greater than any of our $p_i$. However, since $N$ is not a prime, $N$ must be able to be factorized into a product of primes.

Furthermore, the factorization of $N$ must only contain primes of the form 4k+1 since $N$ is odd and all primes of the form 4k-1 are used. This is a contradiction since two numbers which are of the form 4k+1 must have a product of the form 4k+1. Hence, if $N$ was entirely the product of primes of the form 4k+1, $N$ must be a number of the form 4k+1 also. But since $N$ is obviously not expressible as 4k+1, $N$ must itself be a prime that is of the form 4k-1 or have some new prime factor that is of the form 4k-1. Again, by contradiction, there must exist an infinite number of primes of the form 4k-1. The case for 6k-1 is similar. $\blacksquare$

As a student of mathematics, one of the first things I ever learned was the concept of a proof. In fact, the first proof I ever saw was Euclid's classic proof that there were an infinite number of primes.If you have never seen Euclid's proof before, it goes something like this:Recently, I learned that a variant of Euclid’s proof can be used to prove the infinitude of primes of the form 4k-1 and 6k-1. The proof for 4k-1 is below:Amazingly, there is a general theorem which states that a Euclid-style proof can only be used to prove the infinitude of primes of the form $p \equiv a\bmod{m}$, where $a^2 \equiv 1 \bmod{m}$ for any two natural numbers $a$ and $m$ (proven here by Murty, 1988). For example, this means that infinitude any primes of the form $p \equiv -1 \bmod 4$ may be proven (as we did above), but any primes of the form $p \equiv 3 \bmod 7$ cannot be proven by Euclid's proof.I find Euclid’s classic proof of the infinitude of primes beautiful -- not just because it establishes an amazing result in an elegant way, but because of its vast reusability. For these reasons, it is easily my favourite proof.From the theorem above, we may also prove infinitude of primes of the form 4k+1 by a Euclid style proof. Try proving this statement! Be warned though, there is an extra trick to defining $N$ than just multiplying all primes of the form 4k+1 together. The solution can be found in the slides of Greg Martin, as referenced in the first sentence of this post.