Primes not dividing sequence $a_{n+1} = 1 + a_0 a_1 \cdots a_n$
Let me write it here, because I had a typo in the comment.
First observe that
$$
\begin{align}
a_{n+2}&amp;=1+a_0a_1\dotsm a_na_{n+1}\\
&amp;=1+a_0a_1\dotsm a_n(1+a_0a_1\dotsm a_n)
\end{align}$$
So $$4a_{n+2}=(2a_0a_1\dotsm a_n+1)^2+3$$
This implies that for an odd prime $p$ that divides $a_{n+2}$, we must have that $-3$ is a quadratic residue modulo $p$. By quadratic reciprocity, $-3$ is a quadratic residue modulo $p$, when $p$ itself is a quadratic residue modulo $3$. Therefore, any odd prime of the form $3K+2$, with $K\in\mathbb{Z}$ doesn't have $-3$ as quadratic residue and  cannot divide $4a_{n+2}$ for any $n\geq0$.
Since there are infinitely many primes of the form $3K+2$ (see here for example), then an infinite collection of primes that don't divide any of the elements of the sequence are the primes of the form $3K+2$, excluding from them $2$, the primes that divide $a_0$ and the primes that divide $a_1$.