Infinitude of odd primes of the form $4n+3$. I have read in a book proof regarding the infinitude of primes of the form $4n+3$. The issue is there are missing steps (as per the understanding level of mine),
 and want to get the 'filled missing steps' vetted (highlighted by $\color {blue}{\textrm{blue color}}$). Will ask doubts, when am unable to understand text. For comparison, an image of the book's proof is given. 
  It starts with the statement that: Any odd prime is either of the form $4n+1$ or $4n+3$ for suitable $n \in \mathbb{
Z}$. For later requirement, it is shown 'earlier' that product for two odd integers, of the form $4n+1$ is also of the same form, as repeated below: > Let two primes be $4n_1+1, 4n_2+1$ with the product being : $(4n_1+1)(4n_2+1) = 4(4n_1n_2 + n_1+n_2) + 1$, i.e. of the same form.
Next, it shows infinitude of primes for the $4n+3$ form by contradiction approach, by assuming a finite number ($N$) of such primes $p_1,p_2, p_3, ..., p_N$ being of the form $4n+3$ in ascending order, with $p_1=3, p_2=7,$ $p_3=11, p_4=15$, etc.  Hence, the product of such primes (let, $M$) would also be of the same form. Let $$M = 4(p_2p_3...p_N)+3$$


Doubt 1: Why $p_1$ is removed just for the sake of it being divisible by $3$? Will it not affect the value of $M$?


Next, it states that by Theorem 9, $M$ is not divisible by any of $p_2, p_3, ..., p_N$. 


Doubt 2: That proof considers all factors $p_1p_2p_3....p_N$ (i.e. not ignores $p_1$),


takes $M =p_1p_2p_3....p_N+1$, and relies on division of $M$ and $p_1p_2p_3...p_N$ by a (among the N possible) factor, let $p_1$. 
Then shows by contradiction that the left side ($\frac{M}{p_1} - p_2p_3...p_N$ is an integer, while the r.h.s. ($\frac{1}{p_1}$) is not an integer.
Further, $3\nmid M$, as $3\nmid 4$, $\color {blue}{\textrm{as in the equality $M - 4(p_2p_3...p_N) =3$, both terms on the l.h.s.}}$
$\color {blue}{\textrm{need be divisible by $3$.}}$
Hence, $M$ has the form $4n+3$, but no prime factor among $p_2, p_2, p_N$. The alternate form left for prime factors of $M$ is $4n+1$.
But, the product of such prime factors has already been shown to be of the form $4n+1$, rather than the form of $M$, i.e. $4n+3$.
Hence, by contradiction, there are an infinite number of primes of the form $4n+3$. 


Doubt 3: Does the author take that all numbers of the form 4n+3 are odd primes? If so, then should have indicated that. Does he expect the reader to provide a proof of the same? I mean that for me saying that there are infinitely many primes of the form 4n+3, is different from saying that: 'all' numbers of the given form are prime. Also, what about the numbers of the form 4n+1. Are those also same for the primality criteria as 4n+3?


Have found very good links on MSE, here, here, here, here, & also here.

 A: Doubt 1: We are defining $M$ by this expression. There is no reason we can't choose to omit $3$ from the product in the definition.
Doubt 2: Not quite sure I understand what you're saying here. Yes, it is true that $M$ must not be divisible by any of the $3, p_2,\ldots p_N$ since its first term is divsible by $p_2\ldots p_N$ but not by $3$ and its second term is divisble by $3$ but not any of the $p_2\ldots p_N$.
Doubt 3: No the author does not assume all numbers of the form $4n+3$ are prime. Why do yo think they do? They prove there are infinitely many primes of that form by assuming there are finitely many and deriving a contradiction. I don't get what you're asking regarding $4n+1,$ but there are an infinite number of primes of that form as well.
I think you are missing some essential part of the logic here, but I can't tell what it is. Feel free to ask follow-up questions.
A: Doubt 1: The number M is simply defined to be the product of all of the primes of the form 4n+3 except for 3. M doesn't need to be the product of all primes of that form: we simply need an impossible thing to prove by contradiction. M should exist as defined, yet it has impossible properties. 
Doubt 2: $p_2,..,p_N$ are not devisors by the earlier theorem. 3 is not  a divisor since the definition of M that spawned doubt 1 makes it so (i.e 3 is not a factor of $4(p_1...p_n)$, it will not be a factor of the number three larger than this)
I think the doubts come from the assumption that M must include all primes of the form 4n+3 when in fact it is just a number suitable for he proof
