An infinite product...

Let the value of this product be V. $\therefore V = 2^{1/4} \times 2^{2/8} \times 2^{3/16} \dots ∞$

$\therefore V = 2^{1/4 + 2/8 + 3/16 + \dots ∞}$

Let $V = 2^S$

Now, $S = \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \dots ∞$

This is an arithemetico-geometric series.

We have, $\frac{1}{2} (S) = 0 + \frac{1}{8} + \frac{2}{16} + \frac{3}{32} + \dots ∞$

Hence, we get $S - 1/2 (S) = \big( \frac{1}{4} - 0 \big) + \big( \frac{2}{8} - \frac{1}{8} \big) + \big( \frac{3}{16} - \frac{2}{16} \big) + \dots ∞$

$\therefore 1/2 (S) = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots ∞$

This is an infinite geometric series with first term $= \frac{1}{4}$ and common ratio $= \frac{1}{2}$ .

Hence, $1/2 (S) = \frac{1/4}{1-\frac{1}{2}}$

That's why $1/2 (S) = 1/2$

Hence, $S = 1$

And $V = 2^S = 2^1 = \boxed{2}$