[ P=2^{1/2}\cdot 4^{1/4}\cdot 8^{1/8}\cdot 16^{1/16}\cdots ] [ =2^{1/2}\cdot(2^2)^{1/4}\cdot(2^3)^{1/8}\cdot(2^4)^{1/16}\cdots =2^{\sum_{n=1}^{\infty} n/2^n} ]

[ \sum_{n=1}^{\infty}n x^n=\dfrac{x}{(1-x)^2},\quad x=\dfrac12\implies\sum n/2^n=2 ]

[ P=2^2=4 ]

Proved.