Square root of increasing Exponents of 2
Write it as:
$$2\cdot (2^3)^{1/2}\cdot (2^4)^{1/4}\cdot (2^5)^{1/8}...$$
$$2^{1+\frac{3}{2}+\frac{4}{4}+\frac{5}{8}+...}=2^{1+S}$$
Where $$S=\sum_{n=3}^{\infty}\frac{n}{2^{n-2}}=4\sum_{n=3}^{\infty}\frac{n}{2^n}$$
Note that $$\sum_{n=0}^{\infty}\frac{n}{2^n}=2$$ So $$\sum_{n=3}^{\infty}\frac{n}{2^n}=2-\frac{1}{2}-\frac{2}{4}=1$$
Thus $S=4$ and the value is $32$.