What is $2^{\frac{1}{4}}\cdot4^{\frac{1}{8}}\cdot8^{\frac{1}{16}}\cdot16^{\frac{1}{32}}\cdots\infty$ equal to? I came across this question while doing my homework: 

$$\Large 2^{\frac{1}{4}}\cdot4^{\frac{1}{8}}\cdot8^{\frac{1}{16}}\cdot16^{\frac{1}{32}}\cdots\infty=?$$

$$\small\text{OR}$$
$$\large\prod\limits_{x=1}^{\infty} (2x)^{\frac{1}{4x}} = ?$$

My Attempt:
$\large 2^{\frac{1}{4}}\cdot4^{\frac{1}{8}}\cdot8^{\frac{1}{16}}\cdot16^{\frac{1}{32}}\cdots\infty$
$\large \Rightarrow 2^{\frac{1}{4}}\cdot2^{\frac{2}{8}}\cdot2^{\frac{3}{16}}\cdot2^{\frac{4}{32}}\cdots\infty$
$\large \Rightarrow 2^{(\frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \cdots \infty)}$
OR, $\large 2^{\space (\sum\limits_{x=1}^{\infty} \frac{x}{2^{x+1}})}$
$\cdots$ That's it ... I am stuck here ... It does not resemble any series I know...


How do you do it? Thanks!


Olympiad Tricks
\begin{align*}
  y &=
  1^{\frac{1}{2}} \times 2^{\frac{1}{4}} \times 4^{\frac{1}{8}} \times
  \ldots \times 2^{\frac{n-1}{2^{n}}} \times \ldots \\[3pt]
  y^{2} &= 
  2^{\frac{1}{2}} \times 4^{\frac{1}{4}} \times 8^{\frac{1}{8}} \times
  \ldots \times 2^{\frac{n}{2^{n}}} \times \ldots \\[3pt]
  \frac{y^{2}}{y} &=
  \frac{2^{\frac{1}{2}} \times 4^{\frac{1}{4}} \times 8^{\frac{1}{8}} \times
  \ldots \times 2^{\frac{n}{2^{n}}} \times \ldots}
       {1^{\frac{1}{2}} \times 2^{\frac{1}{4}} \times 4^{\frac{1}{8}} \times
  \ldots \times 2^{\frac{n-1}{2^{n}}} \times \ldots} \\[3pt]
  y &=
  2^{\frac{1}{2}} \times 2^{\frac{1}{4}} \times 2^{\frac{1}{8}} \times
  \ldots \times 2^{\frac{1}{2^{n}}} \times \ldots \\[3pt]
  &=2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{n}}+\ldots} \\[3pt]
  &=2^{1} \\[3pt]
  &=2
\end{align*}
