Consider a non-empty set $S$ of primes, with the property that, for every finite subset $S'\subset S$, all the primes dividing $\left(\prod\_{k\in S'}k\right)+1$ are in $S$.


For instance, it can easily be proven that $2\in S$ (if not, then the smallest member $q$ of $S$ is odd, hence, $q+1$ is even, and thus $2\in S$). In fact, with this way, $2+1=3\in S$, $2\cdot 3+1=7\in S$, $2\cdot 7+1=5\in S$, and so on. 


It seems this set must contain all primes, but I could not prove it. 


I tried to use that if $p\_n$ is the smallest prime that is not contained in $S$, then $p\_1,p\_2,\dots,p\_{n-1}\in S$. $p\_n\sim n\log n$ for $n$ large, and if I could somehow show that there is enough residues that can be constructed using subset-products of $p\_1,\dots,p\_{n-1}$ this should have been handled, but how? Can anybody see a way to do it?



A set of prime numbers
If I'm not mistaken, it is true that $S$ must contain all primes.


First of all it is obvious that $S$ is infinite -- indeed, as Euclid teach us, if $S$ is finite then $\left(\prod\limits\_{p\in S} p\right) + 1$ is coprime to all primes in $S$ and therefore has at least one prime factor not in $S$.


Now let $p$ be a prime number. We will show that $p\in S$. Assume by contradiction that $p\notin S$. Let $R = \{r\_1, \ldots , r\_k\}$ be residues modulo $p$ that appear infinitely often as residues of primes from $S$ modulo $p$. Since $S$ is infinite $k \ge 1$. Now let $A = \{a\_1, \ldots, a\_n\}$ be all elements of subgroup of $(\mathbb{Z}/p\mathbb{Z})^\*$, spanned by all $r\_i$. Also denote by $b$ product of all other primes from $S$ whose residues modulo $p$ are not from $R$.


Note that we can form each $a\_i$ as product of different primes from $S$ (and moreover we can use only primes congruent to numbers from $R$ modulo $p$). Then we can also have $ba\_i$.


If some of $ba\_i + 1$ is congruent to $0$ mod $p$ then we found $p$ and we are done. Thus we can assume that $ba\_i + 1$ is never congruent to $0$. Also note that $ba\_i$ never congruent to $0$ (since $p\notin S$ by assumption) and all $ba\_i$ are pairwise distinct(since $p$ is prime). Therefore $ba\_i + 1$ is never congruent to $1$. Finaly note that $1\in A$. From all this we can deduce that there is some $i$ such that $ba\_i + 1\notin A$. Let $C$ be corresponding number. Let $q$ be any of its prime divisors. Note that $q$ can't be from exceptional set since we added all of them through $b$. Thus $q$ is congruent to some $r\_i$. But then $ba\_i + 1\in A$, as product of primes from $R$ -- a contradiction.


