Observe that $a_{n+1} - 2 = a_n^2(a_n - 2)$ for all $n \ge 1$. By induction on $n$ we obtain $a_{n+1} - 2 = 3a_n^2a_{n-1}^2 \cdots a_1^2$ for all $n \ge 1$. Therefore $a_{2011} + 1 = 3(a_{2010}^2a_{2009}^2 \cdots a_1^2 + 1) = (a_{2010}a_{2009} \cdots a_1)^2 + 1$.

Let $p \equiv 3 \pmod 4$ be a prime divisor of $a_{2011} + 1$. It is well known that if $q$ is a prime divisor of $(a_{2010}a_{2009} \cdots a_1)^2 + 1$, then $q \equiv 1 \pmod 4$ or $q = 2$. Thus $p|3$. That is $p = 3$.