Consider $$\prod_{i=1}^n(2^i+1)=(2+1)(2^2+1)(2^3+1)\dots(2^n+1)$$ I am looking for upper and lower bounds on this expression.
A simple lower bound is $$ \prod_{i=1}^n(2^i+1)\geq \prod_{i=1}^n(2^i)=2^{n(n+1)/2}$$ which just takes the leading term.
Edit: As suggested in the comments, we can rewrite the product as $$ \prod_{i=1}^n(2^i+1)= 2^{n(n+1)/2} \times \prod_{i=1}^n(1+ 1/2^i)$$ with the product on the RHS being studied extensively in the linked answer and others posted in the comments. It turns out that $$ \prod_{i=1}^{n=\infty}(1+ 1/2^i) \approx 2.384231029 $$
I can do a bit better by considering all products where I select a single 1 $$(1)(2^2)\dots\dots(2^n),\\ (2^1)(1)(2^3)\dots(2^n),\\(2^1)(2^2)(1) (2^4)\dots(2^n),\\\dots$$ This gives $$ \prod_{i=1}^n(2^i+1) \geq \left(\sum_{k=0}^n2^{-k}\right) \times \prod_{i=1}^n(2^i) = (2-2^{-n})\times2^{n(n+1)/2}$$ Now, I could go on to consider products where one selects two 1s, three 1s and so on. Here, my combinatorics shuts down... Also apparently my expression is related to the $q$-Pochhammer symbol $(a,q)_n$ as $$\prod_{i=1}^n(2^i+1)= 1/2 \times (-1,2)_{n+1}$$. But searching for asymptotics or bounds didnt help me much yet.
I am looking for the best lower bound you can get. In particular is $$\prod_{i=1}^n(2^i+1) \geq \tfrac{5}{2} \times 2^{n(n+1)/2}\; ???$$ Also, apparently the following upper bound holds $$\prod_{i=1}^n(2^i+1)\leq 3 \times 2^{n(n+1)/2}$$ How to show that one?