What is wrong with this proposed proof of the twin prime conjecture?

I was thinking on the twin prime conjecture, that there are an infinite number of twin primes... I came up with a proof. I have to think that it is incomplete or wrong, because many great minds have thought on this previously. I can't see the issue, so I thought I would raise it in a broader forum. What is wrong with this proof?

1) A number n is prime iff n mod p is non zero for every prime number 1 < p < n

This is easy to prove from the definition of a prime number and mod. It just says that n is not equally divisible by another prime, making it prime.

Let $p_n$ = the nth prime $p_0 = 1$, $p_1 = 2$ ...

Consider $N_n = \Pi_0^n p_n$.

It is easy to see that $N_n \mod p_j = 0 $ for all primes $0 < j < n$

$(N_n + 1) \mod p_j = 1 $ for all primes $0 < j < n$, and therefore, from #1 must be prime

$(N_n - 1) \mod p_j = (p_j - 1) $ for primes $0 < j < n$, and therefore, from #1 must be prime as well

The set $p_n$ has infinite members (as shown by Euclid) , so there are infinite $N_n$. Therefore there are an infinite set of twin prime numbers $(N_n-1,N_n+1)$ .

This repeats a common misconception about Euclid's proof. Your argument does not show that either $N_n+1$ or $N_n-1$ is prime, but rather that these numbers must be divisible by a prime greater than $p_n$. Indeed, $N_4=210$ has $N_4+1$ prime but $N_4-1=209$ is divisible by $11$.

$$N_4-1=p_0p_1p_2p_3p_4-1=1\cdot2\cdot3\cdot5\cdot7-1=210-1=209=11\cdot19$$ $$N_6+1=p_0p_1p_2p_3p_4p_5p_6+1=1\cdot2\cdot3\cdot5\cdot7\cdot11\cdot13+1=30030+1=30031=59\cdot509$$

$N_7+1 = 510511 = 19 ⋅ 97 ⋅ 277$, $N_7-1 = 510509 = 61 ⋅ 8369$ is the first example where both numbers are not primes. I would suggest that not only is $(N_k-1, N_k+1)$ not always a twin prime pair, but that this would actually be quite rare.

According to http://primes.utm.edu/top20/page.php?id=5 $N_k+1$, $N_k-1$ have been tested for k ≤ 100,000 with very few primes found, and with no twin primes found beyond the pair (2309, 2311).