Number of factors of $2^{p_1\cdots p_n}+1$

Question

Let $p_1 <p_2 \cdots <p_n$ be the ordered prime numbers starting with $p_1=5$.

Let $a_n = 2^{p_1\cdots p_n}+1$.

It is elementary to show that $a_n$ has at least $4^n$ factors but this bound ($\tau(a_n)\geq 4^n$) does not seem to be sharp, hence my questions:

1. Do we know better bounds than $4^n$?
2. Could we estimate $\tau(a_n)$ when $n\to \infty$?

EDIT Nov 9, 2019 With the link I mentioned in comment I was able to show that $\tau(a_n)\geq 2^{2^{n-1}}$.

Answer 1

$\tau(2^{p_1\cdots p_n}+1) \ge 2^{2^n}$ is easy using Zsigmondy's theorem. For every $d\big|p_1\cdots p_n$, except for $d=3$, there is a primitive prime divisor of $2^d+1$. Hence, $2^{p_1\cdots p_n}+1$ has at least $2^n$ distinct prime divisors.