Easier guessed than proven

Let us compute the first few terms of this sequence:

$[2, 3, 7, 43, 139, 50207, \dots]$

I claim that $5$ is the smallest prime that will never be a member of this sequence. What does a 5 in this sequence imply?

$p_n = p_1p_2\dots p_{n-1} + 1 \equiv 0 \pmod{5}$

And not only this, but 5 would be the largest prime divisor of $p_n$ , because of how the sequence is defined. The only other possible prime divisors are 2 and 3, but because $p_1 = 2$ and $p_2 = 3$ , we must have that $p_n \not\equiv 0 \pmod{2, 3}$ . Our number must be a power of 5. Now we will substitute our new information. For some $k \in \mathbb{N}$ ,

$p_1p_2\dots p_{n-1} + 1 = 5^k \:\:\Rightarrow$

$p_1p_2\dots p_{n-1} = 5^k - 1 = 4(5^{k-1} - 5^{k-2} + 5^{k-3} - \dots - 5 + 1)$

We can see that the RHS is divisible by $p_1 = 2$ , but twice. From this it follows that if we can prove that 2 will not appear again in the sequence, then we will have a contradiction and prove our claim. By similar reasoning, we can do the same thing as before. For some $k \in \mathbb{N}$ ,

$p_1p_2\dots p_{n-1} = 2^k - 1$

The LHS is divisible by $p_1 = 2$ , yet the RHS is odd. This is a contradiction, and thus 2 will not appear twice, and then it directly follows that 5 is the smallest prime that will never appear in the defined sequence.