Can $(1-\frac{1}{2})(1-\frac{1}{2^2})(1-\frac{1}{2^3})...(1-\frac{1}{2^{n-1}})(\frac{1}{2^n})$ be simplified? Can $(1-\frac{1}{2})(1-\frac{1}{2^2})(1-\frac{1}{2^3})...(1-\frac{1}{2^{n-1}})(\frac{1}{2^n})$ be simplified? It seems like an expression from a simple induction proof problem that's missing its result.
 A: You could write
$$\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{7}{8} \cdot \frac{15}{16} \cdot \dots \cdot \frac{2^{n - 1} - 1}{2^{n - 1}} \cdot \frac{1}{2^n}$$
and then the denominator is
$$2 \cdot 4 \cdot \dots 2^n = 2^{1 + 2 + 3 + \dots + n} = 2^{n(n + 1)/2}$$
while the numerator is
$$(2 - 1) (2^2 - 1) (2^3 - 1) \cdots (2^{n - 1} - 1) = \prod_{k = 1}^n (2^k - 1) = (-1)^n (2; 2)_n$$
where $(2; 2)_n$ is a $q$-Pochhammer symbol. So the final "simplification" would be
$$\frac{(-1)^n (2; 2)_n}{2^{n(n + 1)/2}}$$
A: Still no closed form, just another reformulation.
If I use the logarithm of your product $p_n$ (omitting the trailing $1/2^n$-factor), express the logarithms by their Mercator series, and change order of evaluation of the double sums then I get the following sum 
$$ \lim_{n \to \infty } \log(p_n) = - \sum_{k=1}^n {1 \over 2^k-1 }\frac 1k $$  
I think I've read in some article of Euler, that for the sum (without the 1/k-cofactor) "we cannot find a simplification" 
A: It can be transformed into
$$
\frac{(2-1)(2^2-1)(2^3-1)\cdots(2^{n-1}-1)}{2^{1+2+3+\cdots +n}}=
\frac{(2-1)(2^2-1)(2^3-1)\cdots(2^{n-1}-1)}{2^{n(n+1)/2}}
$$
but whether this is a simplification I don't know.
