Infinitely number of primes in the form $4n+1$ proof

Suppose $n>1$ is an integer. We define $N=(n!)^2 +1$. Suppose $p$ is the smallest prime divisor of $N$. Since $N$ is odd, $p$ cannot be equal to $2$. It is clear that $p$ is bigger than $n$ (otherwise $p \mid 1$). If we show that $p$ is of the form $4k+1$ then we can repeat the procedure replacing $n$ with $p$ and we produce an infinite sequence of primes of the form $4k+1$.

We know that $p$ has the form $4k+1$ or $4k+3$. Since $p\mid N$ we have $$ (n!)^2 \equiv -1 \ \ (p) \ . $$ Therefore $$ (n!)^{p-1} \equiv (-1)^{ \frac{p-1}{2} } \ \ (p) \ . $$ Using Fermat's little Theorem we get $$ (-1)^{ \frac{p-1}{2} } \equiv 1 \ \ (p) \ . $$ If $p$ was of the form $4k+3$ then $\frac{p-1}{2} =2k+1$ is odd and therefore we obtain $-1 \equiv 1 \ \ (p)$ or $p \mid 2$ which is a contradiction since $p$ is odd.

There's indeed another elementary approach:

For every even $n$, all prime divisors of $n^2+1$ are $ \equiv 1 \mod 4$. This is because any $p\mid n^2+1$ fulfills $n^2 \equiv -1 \mod p$ and therefore $\left( \frac{-1}{p}\right) =1$, which is, since $p$ must be odd, equivalent to $p \equiv 1 \mod 4$.

Assume that there are only $k$ primes $p_1,...,p_k$ of the form $4m+1$. Then you can derive a contradiction from considering the prime factors of $(2p_1...p_k)^2+1$.

(There's also an elementary approach to show that there are infinitely many primes congruent to $1$ modulo $n$ for every $n$, but that one gets rather tedious. (See: Wikipedia as a reference.))

There are infinite primes in both the arithmetic progressions $4k+1$ and $4k-1$.
Euclid's proof of the infinitude of primes can be easily modified to prove the existence of infinite primes of the form $4k-1$.

Sketch of proof: assume that the set of these primes is finite, given by $\{p_1=3,p_2=7,\ldots,p_k\}$, and consider the huge number $M=p_1^2 p_2^2\cdots p_k^2-1$. $M$ is a number of the form $4k-1$, hence by the fundamental theorem of Arithmetics it has a prime divisor of the same form. But $\gcd(M,p_j)=1$ for any $j\in[1,k]$, hence we have a contradiction.

Given that there are infinite primes in the AP $4k-1$, is that possible that there are just a finite number of primes in the AP $4k+1$? It does not look as reasonable, and indeed it does not occur. Let us define, for any $n\in\mathbb{N}^+$, $\chi_4(n)$ as $1$ if $n=4k+1$, as $-1$ if $n=4k-1$, as $0$ if $n$ is even. $\chi_4(n)$ is a periodic and multiplicative function (a Dirichlet character) associated with the $L$-function $$ L(\chi_4,s)=\sum_{n\geq 1}\frac{\chi_4(n)}{n^s}=\!\!\!\!\prod_{p\equiv 1\!\!\pmod{4}}\left(1-\frac{1}{p^s}\right)^{-1}\prod_{p\equiv 3\!\!\pmod{4}}\left(1+\frac{1}{p^s}\right)^{-1}. $$ The last equality follows from Euler's product, which allows us to state $$ L(\chi_4,s)=\prod_{p}\left(1+\frac{1}{p^s}\right)^{-1}\prod_{p\equiv 1\!\!\pmod{4}}\frac{p^s+1}{p^s-1}=\frac{\zeta(2s)}{\zeta(s)}\prod_{p\equiv 1\!\!\pmod{4}}\frac{p^s+1}{p^s-1}. $$ If the primes in the AP $4k+1$ were finite, the limit of the RHS as $s\to 1^+$ would be $0$.
On the other hand, $$ \lim_{s\to 1^+}L(\chi_4,s)=\sum_{n\geq 0}\frac{(-1)^n}{2n+1}=\int_{0}^{1}\sum_{n\geq 0}(-1)^n x^{2n}\,dx=\int_{0}^{1}\frac{dx}{1+x^2}=\frac{\pi}{4}\color{red}{\neq} 0 $$ so there have to be infinite primes of the form $4k+1$, too.

With minor adjustments, the same approach shows that there are infinite primes in both the APs $6k-1$ and $6k+1$. I have just sketched a simplified version of Dirichlet's theorem for primes in APs.