Roots and just roots!

Let $\begin{aligned}\text{P} & =2\times\sqrt{2}\times\sqrt[3]{\sqrt{2}}\times\sqrt[4]{\sqrt[3]{\sqrt{2}}}\cdots \\ & = 2^{1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\cdots} \\ & = 2^{1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots} \\ & = 2^{\left(\displaystyle\sum_{n=0}^{\infty}\frac{1}{n!}-1\right)} \\ &= 2^{(e-1)}\end{aligned}$

Thus $e-1 \approx\boxed{1.718281}$ .