Quickly show that $\Bigg|(-1/2)^{(-1/2)^{(-1/2)}}\Bigg|=e^{\pi\sqrt{2}}$ 
Question: Is it true and can we quickly show that $$\Bigg|(-1/2)^{(-1/2)^{(-1/2)}}\Bigg|=e^{\pi\sqrt{2}}$$

Here was my solution. Abusively I write $a/b$ for the fraction ${a \above 1.5pt b} $. I write $i=\sqrt{-1}$. If $z$ is any complex number I denote the absolute value of $z$ by $|z|$.  

Solution: We start by computing the tower top down. Observe $$(-1/2)^{(-1/2)}= {1\above 1.5pt \sqrt{{-1\above 1.5pt 2}}}={1 \above 1.5pt i \sqrt{1 \above 1.5pt 2}}={1 \above 1.5pt i {\sqrt{1} \above 1.5pt \sqrt{2}}}={\sqrt{2}\above 1.5pt i\sqrt{1}}={\sqrt{2} \above 1.5pt i}$$ I multiply the numerator and denominator of the the RHS of the above equation by the conjugate of the denominator of ${\sqrt{2}\above 1.5pt i}$ and simplify; $${\sqrt{2}(-i) \above 1.5pt i(-i)}={-i\sqrt{2}\above 1.5pt -(-1)}=-i\sqrt{2}$$ Now I can compute the base of the tower $$\left(-{1 \above 1.5pt 2}\right)^{-i\sqrt{2}}=(-1)^{-i\sqrt{2}}(2)^{-i\sqrt{2}}$$ I bring Euler's identity into play and rewrite $(-1)^{-i}$ as $e^{\pi}$. I also recall that absolute value is multiplicative for real and complex numbers. In particular.
  $$\Bigg|(-1)^{-i\sqrt{2}}(2)^{-i\sqrt{2}}\Bigg|=\Bigg|(-1^{-i})^{\sqrt{2}}(2^{-i})^{\sqrt{2}}\Bigg|=\Bigg|e^{\pi\sqrt{2}} \Bigg| \times\Bigg|{\overline{2^{-i}} \above 1.5pt 2^i}\Bigg|=e^{\pi\sqrt{2}}$$ This completes the solution.

Is there a quicker way to show this ?  
The motivation here were cheap, possibly trivial, ways to rewrite $e^{\pi\sqrt{2}}$, using only finitely many combinations of rational numbers or integers in the rewrite. Here are a few more examples:


*

*$\Bigg|(-1/4)^{(-1/4)^{(-1/4)}}\Bigg| = {e^{\pi} \above 1.5pt 4}$ in
particular $\Bigg|(-1)^{(-1/4)^{(-1/4)}}\Bigg| = e^{\pi}.$ See
Gelfond's
constant

*$\Bigg|(-2)^{(1/4)^{(-1/4)}}\Bigg| = 2^{\sqrt{2}}.$ See Gelfond Schneider Constant.


The rewrites are not unique. 
 A: Since you are multiplying/exponentiating, it makes sense to keep the numbers in polar coordinates. so let $x = -\frac{1}{2} = \frac{1}{2}e^{i\pi} = e^{-\log 2 + i\pi}$. Then
\begin{align}
x^{x^x} = \exp(x^x\log x) &=\exp( e^{x\log x}\log x) \\
&= \exp(e^{-\frac{1}{2}(-\log 2 + i\pi)}(-\log 2 +i\pi)) \\
&= \exp(-i\sqrt{2}(-\log 2 + i\pi))\\
&=\exp(\sqrt{2}\pi +i\sqrt{2}\log 2)
\end{align}
Hence $\big|x^{x^x}\big| = e^{\sqrt{2}\pi}$
A: You need to be clear about how you define $a^b$ when $a$ is a negative real number and $b\in\mathbb{C}$.  One standard way is to stipulate
$$a^b=e^{b(\ln|a|+i\pi)}$$
where 
$$e^z=\sum_{n=0}^\infty{z^n\over n!}\quad\text{for } z\in\mathbb{C}\qquad\text{and}\qquad\ln x=\int_1^x{du\over u}\quad\text{for }x\in\mathbb{R^+}$$
With that convention,
$$(-1/2)^{(-1/2)}=e^{(-1/2)(-\ln2+i\pi)}=e^{\ln(\sqrt2)}(\cos(-\pi/2)+i\sin(-\pi/2))=-i\sqrt2$$
and thus
$$(-1/2)^{(-1/2)^{(-1/2)}}=(-1/2)^{-i\sqrt2}=e^{-i\sqrt2(-\ln2+i\pi)}=e^{\pi\sqrt2}e^{i\sqrt2\ln2}=e^{\pi\sqrt2}(\cos(\sqrt2\ln2)+i\sin(\sqrt2\ln2))\approx47.3396584233+70.6208560994i$$
If all you want is the absolute value, then
$$\left|(-1/2)^{(-1/2)^{(-1/2)}}\,\,\,\right|=e^{\pi\sqrt2}\approx85.0196952232$$
