I am a highschool student going through the elementary number theory found the following please help me understanding this.
prove that there are infinitely many primes of the form $4k+3$
My book start this way :
Let $n=4k+3$ and $S=\{d\,|\,d\text{ is a divisor of $n$ of the form $4m+3$}\}$. As $n$ belongs to $S$ so $S$ is not the empty set, so there must be a least element say $p$. Then $p=\min\{d\mid d\in S\}$. Now there are two possibilities that $p$ may be prime or composite, say if $p$ is composite then $p=ab$ as $p=4m+3$ at least one of the $a,b$ must be in the same form. Say $a=4l+3$ for some $l$. Hence as $a<p$ it contradicts our assumption of minimum element. Thus $p$ is prime.
Up to this I got it but after this I am unable to grasp :
So from this we can conclude that every number in the form $4k+3$ must have at least one prime divisor of the same form. Now consider $4(n!-1)+3=4n!-1$ so from the above it should have prime divisor of the same form. But $p$ does not divide $4n!-1$ for any natural number $p$ in $[2,n]$ so $p>n$. Hence there are infinitely many primes.
Please help me in the last part, I did not get why $p$ does not divide $4n!-1$ when $p$ is less than equal to $n$
The missing part of the reasoning is:
Assume there are finitely many primes of the form $4k+3$. We can define $n$ as the highest such prime.
Now consider $n'=4(n!-1)+3=4(n!)-1$. That $n'$ is of the form $4m'+3$, thus we can define $p$ as the least positive divisor of $4(n!)−1$ of the form $4m+3$, and we have seen that such $p$ must be prime.
If it held that $p\le n$, then $p$ would divide $n!$, thus would divide $4(n!)$, thus would not divide $4(n!)−1$. It follows that $p$ is larger than $n$. By construction, it is also of the form $4m+3$, and is prime. That contradicts the assumption in bold.
That assumption is thus false, Q.E.D.