無限に掛けるとどうなる？

\begin{align*} 2^{\frac{1}{4}}4^{\frac{1}{8}}8^{\frac{1}{16}}\cdots & =\prod_{k=1}^{\infty}\left(2^{k}\right)^{\frac{1}{2^{k+1}}}\\ & =\prod_{k=1}^{\infty}\pow\left(2,\frac{k}{2^{k+1}}\right)\\ & =\pow\left(2,\frac{1}{2}\sum_{k=1}^{\infty}\frac{k}{2^{k}}\right)\\ & =\pow\left(2,-\frac{1}{2}\left[2\sum_{k=1}^{\infty}\frac{d}{da}\frac{1}{a^{k}}\right]_{a=2}\right)\\ & =\pow\left(2,-\frac{1}{2}\left[2\frac{d}{da}\frac{\frac{1}{a}}{1-\frac{1}{a}}\right]_{a=2}\right)\\ & =\pow\left(2,-\frac{1}{2}\left[2\frac{d}{da}\frac{1}{a-1}\right]_{a=2}\right)\\ & =\pow\left(2,\frac{1}{2}\left[2\frac{1}{\left(a-1\right)^{2}}\right]_{a=2}\right)\\ & =2 \end{align*}