The product $$2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 16^{\frac{1}{128}} \cdot \ldots$$ to $$\infty$$ is equal to:

We begin by writing every factor in the infinite product with the same base, namely the base $$2$$. In each factor the base is already a power of $$2$$, so we only have to rewrite the exponents.

The given product is

$$ 2^{\frac14}\;\cdot\;4^{\frac1{16}}\;\cdot\;8^{\frac1{48}}\;\cdot\;16^{\frac1{128}}\;\cdot\;\ldots $$

Observe that the bases are consecutive powers of $$2$$:

$$ 2 = 2^{1},\qquad 4 = 2^{2},\qquad 8 = 2^{3},\qquad 16 = 2^{4},\qquad\ldots $$

Let the general term be the one in which the base is $$2^{k}$$. We now look for a general rule for the corresponding exponent.

From the four visible terms we read

$$ \begin{aligned} k &= 1 &\Longrightarrow&\quad \text{exponent}=\frac14,\\[2mm] k &= 2 &\Longrightarrow&\quad \text{exponent}=\frac1{16},\\[2mm] k &= 3 &\Longrightarrow&\quad \text{exponent}=\frac1{48},\\[2mm] k &= 4 &\Longrightarrow&\quad \text{exponent}=\frac1{128}. \end{aligned} $$

Notice that the denominators follow the sequence

$$ 4,\;16,\;48,\;128,\;\ldots $$

Dividing each of these denominators by $$2^{\,k+1}$$ we get

$$ \frac{4}{2^{1+1}} \;=\; 1,\qquad \frac{16}{2^{2+1}}\;=\;\frac12,\qquad \frac{48}{2^{3+1}}\;=\;\frac13,\qquad \frac{128}{2^{4+1}}\;=\;\frac14. $$

Hence the pattern is

$$ \text{exponent for }2^{k} \;=\;\frac1{k\,2^{\,k+1}}. $$

Therefore the k-th factor in the product can be written as

$$ \bigl(2^{k}\bigr)^{\frac1{k\,2^{\,k+1}}} \;=\;2^{\,\frac{k}{k\,2^{\,k+1}}} \;=\;2^{\frac1{2^{\,k+1}}}. $$

So the entire product becomes

$$ \prod_{k=1}^{\infty}2^{\frac1{2^{\,k+1}}} \;=\; 2^{\;\displaystyle\sum_{k=1}^{\infty}\frac1{2^{\,k+1}}}. $$

The exponent is an infinite geometric series. First we recall the formula for the sum of a geometric series:

$$ \text{If }\;|r|<1,\quad \sum_{n=0}^{\infty} r^{\,n}=\frac1{1-r}. $$

Here the common ratio is $$\dfrac12$$. We need the series starting from the term $$n=2$$, because our summation index k begins with $$1$$ and the power of $$2$$ in the denominator is $$k+1$$. Thus

$$ \sum_{k=1}^{\infty}\frac1{2^{\,k+1}} \;=\; \sum_{n=2}^{\infty}\frac1{2^{n}} \;=\; \Bigl(\sum_{n=0}^{\infty}\frac1{2^{n}}\Bigr) - \frac1{2^{\,0}} - \frac1{2^{\,1}} \;=\; \frac1{1-\frac12}\;-\;1\;-\;\frac12 \;=\; 2\;-\;1\;-\;\frac12 \;=\; \frac12. $$

Consequently

$$ 2^{\;\displaystyle\sum_{k=1}^{\infty}\frac1{2^{\,k+1}}} \;=\; 2^{\,\frac12} \;=\; \sqrt2. $$

This evaluates the infinite product to $$\sqrt2 = 2^{\frac12}$$.

Among the given options, this is Option A.

Hence, the correct answer is Option A.