Infinitely many primes of the form $4n+3$ I've found at least 3 other posts$^*$ regarding this theorem, but the posts don't address the issues that I have.
Below is a proof that for infinitely many primes of the form $4n+3$, there's a few questions I have in the proof which I'll mark accordingly.
Proof: Suppose there were only finitely many primes $p_1,\dots, p_k$, which are of the form $4n+3$. Let $N = 4p_1\cdots p_k - 1$. This number is of the form $4n+3$ and is also not prime as it is larger than all the possible primes of the same form. Therefore, it is divisible by a prime $ \color{green}{ \text{(How did they get to this conclusion?)}}$. However, none of the $p_1,\dots, p_k$ divide $N$. So every prime which divides $N$ must be of the form $4n+1$ $ \color{green}{ \text{(Why must it be of this form?)}}$. But notice any two numbers of the form $4n+1$ form a product of the same form, which contradicts the definition of $N$. Contradiction. $\square$
Then as a follow-up question, the text asks "Why does a proof of this flavor fail for primes of the form $4n+1$? $ \color{green}{ \text{(This is my last question.)}}$

$^*$One involves congruences, which I haven't learned yet. The other is a solution-verification type question. The last one makes use of a lemma that is actually one of my questions, but wasn't a question in that post.

Question 1: Any integer is divisible by a prime (you actually don't even need the full strength of the unique factorization theorem; this is provable by induction). 
Question 2: It's not divisible by any of the primes of the form $4n+3$ (namely, $p_1,\ldots,p_k$, which we're assuming is all of them), and it's not divisible by $2$ because it's odd. Thus, any prime dividing it must be of the form $4n+1$.
Question 3: Multiplying an even number of things of the form $4n+3$ together gives something of the form $4n+1$.
