TITLE: On Euclid's proof of the infinitude of primes and generating primes
QUESTION [19 upvotes]: So looking at Euclid's proof he says 
1)take a finite family of primes (F) 
2)multiply all the primes and add one
3)this new number has at least 1 new prime factor
So I was wondering about what kind of primes you get by recursively feeding this process into it self.
Since the number you must factor grows exponentially, it's hard to get a lot of numerical evidence for what happens.
I calculated a few:
[2]-> [2,3]-> [2,3,7]->[2,3,7,43]->[2,3,7,43,13,139]->[2,3,7,43,13,139,3263443]
->[2,3,7,43,13,139,3263443,547,607,1033,31051]-> cannot factor 113423713055421844361000443
[5] (x5)-> [5,2,3,31,7,19,37,3343,79,193662529] -> cannot factor 234069798025176583891
Obviously quite a few primes are missing, 5,11,19,etc from the first list, but could show up later.
So my question is does a finite family of primes exist that eventually generates all the primes?  I figure this probably doesn't have an easy answer, but any information related to this process would be appreciated, or even why it can't be done.

REPLY [7 votes]: Going with what Qiaochu Yuan said about f(x), it follows that we will never get those primes unless we start, even if we don't include multiplicities.  Since we're starting with 'n', then we're taking the prime factors of 'n+1', then we're taking the prime factors of f(n+1), then f(f(n+1)), then f(f(f(n+1))) etc, even if we get, say, a 72 in there, our number is [f(f(f(n+1)))] / 7, which then goes into f(x).
So no, unless you start with the infinite product $p_1 = \prod_{n=1}^\infty 6n-1$, you will never get any of those numbers.
It's funny, though; I'd had a whole demonstration started to show that you'll never get a multiplicity when taking f(f(...f(2)...)), but this is simpler.
As for the Euclid-Mullin sequence, I have no idea.