Sum and Product of Infinite Radicals How to compute $$\sqrt{(1\sqrt{(2\sqrt{(3\dots)})})}$$ &amp; $$\sqrt{(1+\sqrt{(2+\sqrt{(3+\cdots)})})}$$?

I understand that  $$\sqrt{(1\sqrt{(2\sqrt{(3\dots)})})}=(1^{1/2})(2^{1/4})(3^{1/8})\cdots$$

and

$$\sqrt{(1+\sqrt{(2+\sqrt{(3+\cdots)})})}=(1+(2+(3+(\cdots))^{1/8})^{1/4})^{1/2} $$

How to Proceed further?
$$\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\cdots}}}}=C$$

Where $C \approx 1.7579$ (OEIS A072449) is the nested radical constant.  No closed form of this constant is known.



$$\sqrt{1\sqrt{2\sqrt{3\sqrt{4\cdots}}}}=\sigma$$

where $\sigma \approx 1.6616$ (OEIS A112302) is Somos's Quadratic Recurrence Constant.

$\sigma$ has an alternate form (I hesitate to call it a "closed form") of 
$$\sigma =
\exp\left[-2^n \frac{\partial \operatorname{Li_n\left(\frac{1}{2}\right)}}{\partial n}\bigg|_{n=0}+\frac{1}{2} \frac{\partial \Phi\left(\frac{1}{2},s,n+1\right)}{\partial s}\bigg|_{s=0}\right]$$

where $\operatorname{Li}_n(z) = \sum^\infty_{n=1} \frac{z^k}{k^n}$ is the polylogarithm and $\Phi(z,s,a)=\sum_{k=0}^\infty \frac{z^k}{(a+k)^s}$ is the Lerch transcendent, and 

$$\sigma = \exp\left[\int_0^1 \frac{1-x}{(x-2)\log x}\, \mathrm{d}x\right]$$
Let $A=(1^{1/2})(2^{1/4})(3^{1/8})...\implies \log A= \sum_{i=1}^{\infty}\frac{\log i}{2^i}$ and this sum converges to $\approx 0.507834$

http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427em8adj6o2qa

$\implies (1^{1/2})(2^{1/4})(3^{1/8})...$ converges to $\approx e^{0.507834}$.