Prove that there exists infinitely many prime numbers of the form 4k+3 for some integer k. We shall prove this theorem by contradiction. Assume $p_n$ is the largest prime number of the form 4$k_n$+3 for some n $\in$ $\mathbb{N}$. Consider an odd value of n. Even if n is even, we could miss one of the prime numbers to get an odd n. Now, consider
\begin{align*}
N &= 2(4 k_1 +3)(4 k_2 +3)(4k_3+3)...(4 k_n +3) + 1 \\
  &= 4m + 2\cdot 3^n + 1 \\ 
  &= 4m + 2\cdot (4-1)^n + 1 \\
  &= 4m + 2\cdot 4a -2 + 1 \\
  &= 4m + 2\cdot 4a -1 \\
  &= 4m + 3 
\end{align*}
for some m,a $\in$ $\mathbb{N}$
Two cases are possible. N is definitely not divisible by 4$k_i$+3 i $\in$ [1,n].

*

*All numbers between 4$k_n$+3 and N are composite $\implies$ they are formed from a unique prime factorisation of the finite set of prime numbers. Then N is prime. so contradiction.

*There is a new prime between 4$k_n$+3 and N of the form 4k+3 so N is a multiple of that prime(Note that N cannot be a multiple of 4k+1 since we have 2 occurring only once in our expression for N-1), which any way is again a contradiction.

My prof said that this proof misses some cases but I am unable to figure out which case. Please help. I feel pretty confident of my proof.
 A: We shall prove this theorem by contradiction. Assume $p_n$ is the largest prime number of the form 4$k_n$+3 for some n $\in$ $\mathbb{N}$. Now, consider
\begin{align*}
N &= 4(4 k_1 +3)(4 k_2 +3)(4k_3+3)...(4 k_n +3) - 1 \\
  &=4m+3
\end{align*}
for some m $\in$ $\mathbb{N}$
N is definitely not divisible by any of 4$k_i$+3 i $\in$ [1,n].
Now either N is prime or N has at least one prime factor of the form 4$k$+3 apart from itself, not in this list since if N had all prime factors of the form 4k+1, then N would have been of the form (4$u_1$+1)(4$u_2$+1)(4$u_3$+1)...(4$u_n$+1) = 4l+1 only. so a contradiction. Hence proved.
I accommodated some of Peter's changes, hope this proof is correct.
A: Your self-answer proof has logical gaps; I started editing it, but ended up just rewriting it:

Asssume that there are finitely many primes of the form $4k+3;$ let them be $p_1,p_2,\ldots,p_n.$
Let $N=4p_1 p_2 \ldots p_n-1=4(p_1 p_2 \ldots p_n-1)+3.$ Since $N$ is odd, each of its prime divisors must be of the form $4k+1$ or $4k+3.$
The identity $$(4r+1)(4s+1)=4(4rs+r+s)+1$$ shows that products of numbers of the form $4k+1$ are of the same form. Thus, since $N$ is not of this form, it has a prime divisor not of this form.
Hence, $N$ has a prime divisor $q$ of the form $4k+3,$ which is one of $p_1,p_2,\ldots,p_n.$ Therefore, $$q\mid N\quad \text {and} \quad q\mid p_1 p_2 \ldots p_n\\ q\mid(4p_1 p_2 \ldots p_n-N) \\ q\mid1,$$ i.e., $q$ is not prime. Since this is a contradiction, our assumption is false; thus, there are infinitely many primes of the form $4k+3.$
