Q:

are all the numbers in Euclid proof prime numbers?

in Euclid for infinite primes we define $p_0= 2, p_n = p_0p_1...p_{n-1}+1$ that way we get a series so that for every $n \in \mathbb{N}$ we get that $p_n$ is divided by a prime not one of $ p_0,...,p_{n-1} $. 
my question is: does for every $n \in \mathbb{N}$ we get that  $ p_n $  is a prime itself ?  
i will be grateful if someone will give me a counterexample or a proof for this statement. 

A:

No.
Your series starts by
$$2, 3, 7, 43, 1807, 3263443, 10650056950807, \ldots$$
and $$1807= 13 \times 139$$ $$10650056950807=547 \times 607 \times 1033 \times 3105.$$ Anyway
$$2\times3\times5\times7\times11\times13+1=59 \times 509$$

