Infinite Telescoping Product

We'll look first at the partial product

$\displaystyle\prod_{k=0}^{N} \left(1 + \dfrac{1}{2^{2^{k}}}\right) = \left(1 + \dfrac{1}{2}\right)\left(1 + \dfrac{1}{4}\right)\left(1 + \dfrac{1}{16}\right) .... \left(1 + \dfrac{1}{2^{2^{N}}}\right)$ .

Now multiply this product by $1$ in the form $\dfrac{1 - \dfrac{1}{2}}{1 - \dfrac{1}{2}} = 2*\left(1 - \dfrac{1}{2}\right)$ .

The partial product then becomes

$2*\left(1 - \dfrac{1}{2}\right)\left(1 + \dfrac{1}{2}\right)\left(1 + \dfrac{1}{4}\right)\left(1 + \dfrac{1}{16}\right) ... \left(1 + \dfrac{1}{2^{2^{N}}}\right) =$

$2*\left(1 - \dfrac{1}{4}\right)\left(1 + \dfrac{1}{4}\right)\left(1 + \dfrac{1}{16}\right) .... \left(1 + \dfrac{1}{2^{2^{N}}}\right) =$

$2*\left(1 - \dfrac{1}{16}\right)\left(1 + \dfrac{1}{16}\right) .... \left(1 + \dfrac{1}{2^{2^{N}}}\right) = 2*\left(1 - \dfrac{1}{256}\right) .... \left(1 + \dfrac{1}{2^{2^{N}}}\right)$ .

The "collapsing", or telescoping, of the partial product would continue until we're left with

$2*\left(1 - \dfrac{1}{2^{2^{N}}}\right)\left(1 + \dfrac{1}{2^{2^{N}}}\right) = 2*\left(1 - \dfrac{1}{2^{2^{N+1}}}\right)$ .

As $N \rightarrow \infty$ we have $\dfrac{1}{2^{2^{N+1}}} \rightarrow 0$ , leaving us with an infinite product of $\boxed{2}$ .