How to calculate product $\prod_{k=0}^{n-1}\left (1+\frac{1}{2^{2^k}}\right)$? How can I calculate the following product of series? $$ \prod_{k=0}^{n-1}\left (1+\frac{1}{2^{2^k}}\right)    $$....Can I take a geometric series and compare it with that?
 A: As a hint, consider $n = 3$: 
$\displaystyle\prod_{k = 0}^{2}\left(1+\dfrac{1}{2^{2^k}}\right) = \left(1+\dfrac{1}{2^1}\right)\left(1+\dfrac{1}{2^2}\right)\left(1+\dfrac{1}{2^4}\right)$ 
$= 1  + \dfrac{1}{2^1} + \dfrac{1}{2^2} + \dfrac{1}{2^1} \cdot \dfrac{1}{2^2} + \dfrac{1}{2^4} + \dfrac{1}{2^1} \cdot \dfrac{1}{2^4} + \dfrac{1}{2^2} \cdot \dfrac{1}{2^4} + \dfrac{1}{2^1} \cdot \dfrac{1}{2^2} \cdot \dfrac{1}{2^4}$ 
$= 1 + \dfrac{1}{2^1} + \dfrac{1}{2^2} + \dfrac{1}{2^3} + \dfrac{1}{2^4} + \dfrac{1}{2^5} + \dfrac{1}{2^6} + \dfrac{1}{2^7}$
$= 2-\dfrac{1}{2^7}$
Can you generalize this for larger values of $n$?
A: This is really the same as JimmyK4542's answer, but written in the general case.
If you expand the product, you find that it is the sum of all possible terms of the form
$$\frac{1}{2^{a_0 + a_1 2 + a_2 2^2 + \dots + a_{n-1}2^{n-1}}}$$
where each $a_i$ is either $0$ or $1$. Viewing the exponents as integers written in binary, they run though all integers $j = 0, 1, 2, \dots 2^n - 1$. We therefore find that the product is equal to
$$\sum_{j = 0}^{2^n - 1} \frac{1}{2^j} = 2 - \frac{1}{2^{2^n - 1}}.$$
A: Partial Answer:
If I were you, I would first simplify that to
$$1+\frac{1}{2^{2^k}}=\frac{2^{2^k}+1}{2^{2^k}}$$
and so we have
$$\prod_{k=1}^{n-1}\frac{2^{2^k}+1}{2^{2^k}}$$
By splitting this up, we get
$$\prod_{k=1}^{n-1}\frac{1}{2^{2^k}}*\prod_{k=1}^{n-1}\big(2^{2^k}+1\big)$$
Your insight about there being geometric series is correct, because look what happens when we evaluate the first product:
$$\frac{1}{2^{2^1}}*\frac{1}{2^{2^2}}*...*\frac{1}{2^{2^{n-1}}}$$
$$\frac{1}{2^{2^1+2^2+...+2^{n-1}}}$$
$$\frac{1}{2^{2^n-1}}$$
So we have
$$\frac{1}{2^{2^n-1}}*\prod_{k=1}^{n-1}\big(2^{2^k}+1\big)$$
I'm afraid that's all I've got... I have no idea how to evaluate that last part.
A: Multiply through by $\frac12=1-\frac1{2^{2^0}}$ to get
$$\begin{align}\left(1-\frac1{2^{2^0}}\right)\prod_{k=0}^n\left(1+\frac1{2^{2^k}}\right)&=\left(1-\frac1{2^{2^0}}\right)\left(1+\frac1{2^{2^0}}\right)\prod_{k=1}^n\left(1+\frac1{2^{2^k}}\right)\\&=\left(1-\frac1{2^{2^1}}\right)\left(1+\frac1{2^{2^1}}\right)\prod_{k=2}^n\left(1+\frac1{2^{2^k}}\right)\\&=\left(1-\frac1{2^{2^2}}\right)\left(1+\frac1{2^{2^2}}\right)\prod_{k=3}^n\left(1+\frac1{2^{2^k}}\right)\\&=\left(1-\frac1{2^{2^3}}\right)\left(1+\frac1{2^{2^3}}\right)\prod_{k=4}^n\left(1+\frac1{2^{2^k}}\right)\\&=\quad\vdots\\&=\left(1-\frac1{2^{2^{n+1}}}\right)\end{align}$$

$$\prod_{k=0}^n\left(1+\frac1{2^{2^k}}\right)=2-\frac1{2^{2^{n+1}-1}}$$

