Evaluate $\sqrt[2]{2} \cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots$

$$I=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}+\cdots$$ $$2I=1+1+\frac{3}{4}+\frac{4}{8}+\frac{5}{16}+\frac{6}{32}+\cdots$$ $$2I-I=1+\left(1-\frac 12 \right)+\left(\frac 34 -\frac 24 \right)+\left(\frac 48 -\frac 38 \right)+\left(\frac {5}{16} -\frac {4}{16} \right)+\cdots$$ $$I=1+\frac 12+\frac 14+\frac 18+\cdots=2$$ therefore $$\lim_{n\to \infty }\sqrt[2]{2}\times\sqrt[4]{4}\times\sqrt[8]{8}\times\dots\times\sqrt[2^n]{2^n}=2^2=4$$

I believe this is a "known" representation of the key summation:

$$\sum_{n=1}^{\infty} \frac{n}{2^n} \;=\; \frac{1}{2} + \left( \frac{1}{4} + \frac{1}{4} \right) + \left( \frac{1}{8} + \frac{1}{8} + \frac{1}{8} \right) + \cdots \;=\; 2 $$

I hope this would be useful for you.

So your trying to evaluate $\displaystyle\prod_{n=1}^{\infty}\sqrt[2^{n}]{2^{n}}$. Consider first $\displaystyle\prod_{n=1}^{k}\sqrt[2^{n}]{2^{n}}=2^{\sum_{n=1}^{k}\frac{n}{2^{n}}}$, after developing the product properly. Now you have to make sense of $$\lim_{k\to\infty}\sum_{n=1}^{k}\frac{n}{2^{n}}$$ , so to reach a specific value, note that $\sum_{n=1}^{k}\frac{n}{2^{n}}=\sum_{n=1}^{k}\frac{n-1}{2^{n-1}}-\frac{n}{2^{n}}+\sum_{n=1}^{k}\frac{1}{2^{n-1}}$, therefore

\begin{eqnarray*} \lim_{k\to\infty}\sum_{n=1}^{k}\frac{n}{2^{n}}&=&\lim_{k\to\infty}\sum_{n=1}^{k}\frac{n-1}{2^{n-1}}-\frac{n} {2^{n}}+\sum_{n=1}^{k}\frac{1}{2^{n-1}}\\ &=&\lim_{k\to\infty}-\frac{k}{2^{k}}+\frac{1-\frac{1}{2^{k+1}}}{1-\frac{1}{2}}\\ &=&2 \end{eqnarray*} Finally $$\prod_{n=1}^{\infty}\sqrt[2^{n}]{2^{n}}=\lim_{k\to\infty}\prod_{n=1}^{k}\sqrt[2^{n}]{2^{n}}=\lim_{k\to\infty}2^{\sum_{n=1}^{k}\frac{n}{2^{n}}}=2^{\lim_{k\to\infty}\sum_{n=1}^{k}\frac{n}{2^{n}}}=2^{2}=4$$