Root of powers

$\Large \sqrt{2^{\sqrt{2}}} = (2^{\sqrt{2}})^{1/2} = (2^{1/2})^{\sqrt{2}} = \boxed{\sqrt{2}^{\sqrt{2}}}$ .

$\large\begin{aligned} & N = \sqrt{2^{ \sqrt{2}}} \\& N =\left(2 ^{\sqrt 2 }\right)^{\frac{1}{2}} \\& N =2^{\frac{\sqrt 2}{2}} \phantom{cc}\\& N = \left(2^{\frac{1}{2}}\right)^{\sqrt 2} \\& N = \sqrt{2}^{\sqrt2}\end{aligned}$

$\large\sqrt{2^{\sqrt2}}$

= $\large{{2}^{{1/2}}}^{\sqrt{2}}$

= $\large{\sqrt{2}}^{\sqrt{2}}$ ........................[ $2^{1/2}=\sqrt{2}$ ]

$Let\ a = \sqrt{2} \\ And\ x = \sqrt{2^{\sqrt{2}}} \\x = \sqrt{2^a} \\x^2 = 2^a \\x^2 = (\sqrt{2})^{2a} \\x = \sqrt{2}^a\\x = \sqrt{2}^{\sqrt{2}}$

Look at the image below, to see the solution and/or proofs to this problem, that I got. It's a bit blurry, but you'll live to tell the tale:

I played with this for a while but I couldn't get to any of the answers. So I guessed I must be missing an identity of $\sqrt{2}$ or something. That led me to Gelfond–Schneider constant , which just flat out states $\sqrt{2^{\sqrt{2}}} = \sqrt{2}^{\sqrt{2}}$ but doesn't explain it.

When i calculated the value of both there was slight difference in later digits after decimal. Can anyone explain why that happened?

I think $2^{\sqrt{\frac{1}{2}}}$ is more elegant