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?