TITLE: Constructing prime numbers
QUESTION [26 upvotes]: The classical proof of the infiniteness of prime numbers is to take the $k$ first prime numbers $p_1,\ldots,p_k$, then to form
$$n_k:=1+p_1\cdots p_k.$$
Then $n_k$ has a prime factor, which is none of the $p_j$.
Notice that one could form instead the number
$$n_{k,I}:=\prod_{i\in I}p_i+\prod_{j\not\in I}p_j$$
where $I$ is any subset of $[1,\ldots,k]$ (we have $n_k=n_{k,\emptyset}$), or even
$$m_{k,I}:=\prod_{i\in I}p_i-\prod_{j\not\in I}p_j,$$
provided this number is not $\pm1$.
This suggests the following construction of prime numbers $q_k$ by induction. Start with $k=1$ and $q_1=2$. At step $k$, form the numbers
$$M_{k,I}:=\left|\prod_{i\in I}q_i-\prod_{j\not\in I}q_j\right|.$$
Then $q_{k+1}$ is the least of the prime factors of all these numbers (that is, of their product as $I$ runs over the subsets of $[1,\ldots,k]$). 

Is it true that $q_k=p_k$ for all $k$ ? If not, does the sequence covers all the prime numbers ? If not, how fast does this sequence increases as $k\rightarrow+\infty$ ?

One might enlarge the construction by taking in account the numbers $M_{k,I}$ and
$$N_{k,I}:=\prod_{i\in I}q_i+\prod_{j\not\in I}q_j,$$
but they are way larger.

REPLY [4 votes]: maybe you could take the primes on the products in any powers you want , it is a thought that i had before some years and it is also a natural question that one can make reading Euclid's proof since still the difference of the products is not divisible by any of these primes do you take in this way always the next prime?