Is this statement always true?

If $p_1,…,p_n$ are all primes that exist, than $p_1\cdot p_2\cdot \cdot \cdot p_n + 1$ will always be prime. I am confused because I thought Euclid’s proof of infinity of primes was based on proving existence of a prime that is other than assumed finitely many primes, i.e., $p_1, p_2, …, p_n$. I always thought $p_1\cdot p_2\cdot \cdot \cdot p_n + 1$ was a prime because it is not divisible by any of the prime among $p_1,…,p_n$.

https://primes.utm.edu/notes/proofs/infinite/euclids.html

But this website says that $p_1\cdot p_2\cdot \cdot \cdot p_n + 1$ should not be considered as a prime?

Solutions Collecting From Web of "Is this statement always true?"

$$2\cdot3\cdot5\cdot7\cdot11\cdot13+1=59\cdot509$$

Hint $\ $ Suppose for contradiction that $\,2\,$ and $\,7\,$ are the only primes. Then $\,15 = 2*7+1\,$ has no smaller prime factors so is prime, contradiction. But $15$ is not prime in the real integers. Rather, it is prime only in the hypothetical integers having only the primes $\,2\,$ and $\,7\,$.

Typically one proves that every natural number $n>1$ is divisible by a prime by strong induction. The idea of the proof is to construct a number which is not divisible by any prime supposing that we had only finitely many primes $p_1, \dotsc, p_n$. This number is $p_1p_2\cdots p_n+1$ which leads to a contradiction. If we have infinitely many primes, there could be other primes than $p_1, \dotsc, p_n$ which divide $p_1p_2\cdots p_n+1$.