Understanding the proof of infinitely many primes $\equiv 3\pmod{\!4}$ 
Theorem: There are infinitely many primes congruent to $3 \mod 4$.
Proof: Assume that $p_1 = 3, . . . , p_n$ are primes of the form $p_j ≡ 3 \mod 4$. We will construct a new one by looking at $N = 4p_1 · · · p_n − 1$ (putting $N = 4p_1 · · · p_n+3$ would also work).

There is more to this proof but the rest of it I understand, what I don't get is where it says "$N = 4p_1 · · · p_n − 1$ (putting $N = 4p_1 · · · p_n+3$ would also work)".
I do not understand why we are writing $4p_1 · · · p_n − 1$. Is there a more clear way to prove this theorem?
 A: The key ideas (colored below) are clarified after a slight amount of abstraction. Below we prove a lemma that includes both Euclid's classical proof, as well as the OP (and many others).
Lemma $\ $ Suppose  $\,S\,$ is a set of positive integers that is $\rm\color{#0a0}{closed}$ under multiplication, and $\,\color{#c00}{\bf 1}\in S,\,$ and for any   $\,n\in \Bbb N\,$ there exists a positive integer  $\,c(n)\color{#0af}{\not\in S}\,$ with $\,c(n)\,$ $\rm\color{#90f}{coprime}$ to $\,n.\,$ Then there  exist infinitely many primes not in $\,S.$
Proof $\ $ For induction, let $\,p_1,\ldots p_k\,$ be primes $\,\not\in S.\,$ Then $\,c := c(p_1\!\cdots p_k)\color{#0af}{\not\in S}\,$  so $\,c >\color{#c00}{\bf  1}\,$ hence $\,c\,$ has a prime factor. Not every prime factor of $\,c\,$ lies in $\,S\,$ (else their product $\,c\,$ would be in $\,S\,$ by $\,S\,$ $\rm\color{#0a0}{closed}$ under multiplication). Thus $\,c\,$ has a prime factor $\,p\not\in S.\,$ Since $\,c\,$ is $\rm\color{#90f}{coprime}$ to $\,p_1\cdots p_k\,$ so too is its factor $\,p,\,$ hence $\,p\neq p_i\,$ is a new prime $\not\in S.$

Euclid's proof is the special case $\ S = \{1\}\ $ and $\,\ c(n) = n+1.$
The OP is also a special case: $\, S = 4\,\Bbb N + 1\,$ and $\,c(n) = 4n\!-\!1.\ $ Let's trace this particular proof.
Starting with the empty list of primes with product $= \color{#c00}{\bf 1},$ we construct the new prime
$ 4(\ \ )-1 = 4(\color{#c00}{\bf 1})-1 = 3  =: p_1.\ $ Repeating with the singleton list $\, p_1\,$ leads to
$4(p_1)\!-\!1 = 4(3)\!-\!1 = 11 =: p_2.\ $  Repeating with the list $\, p_1,p_2\,$ leads to
$4(p_1p_2 )-1 = 4(3\cdot 11)-1 =  131 =: p_3.\ $
They remain prime $\ 3, 11, 131, 17291, 298995971 \ $ till we reach the sixth element
$$ n = 89398590973228811 = 8779\cdot 10079\cdot 1010341471$$
where we need to choose a (guaranteed) factor $\,\not \equiv 1\pmod{4},\,$ e.g. the least $= 8779$.
You can find further terms in OEIS sequence A057205
If we use $\,c(n) = 4n\!+\!3\,$ we obtain  $\, 7,31,13,11287,67,\ldots$ (choosing least prime factors)
