Let, $ z\;=\; i^{i^{i^{\ldots \inf}}} \;=\; A+iB \\ \; \\ $ Taking log on both sides,

$ \therefore log \; z \;=\;i^{i^{i^{\ldots \inf}}} log \; i \;=\; z log i \;=\; log(A+iB) \\ \; \\ \; \\ \therefore z \;=\; \dfrac{log \; z}{ log \; i} \\ \; \\ \; \\ \therefore A+iB \;=\; \dfrac{log(A+iB)}{log \; i} \\ \; \\ \; \\ \therefore A+iB \;=\; \dfrac{\dfrac{1}{2} log(A^2+B^2) + itan^{-1} \dfrac{B}{A} }{i \dfrac{\pi}{2} } \;=\; \dfrac{\dfrac{i}{2} log(A^2+B^2) + i^2tan^{-1} \dfrac{B}{A} }{i^2 \dfrac{\pi}{2} } \\ \; \\ \; \\ \dfrac{ tan^{-1} \dfrac{B}{A} }{ \dfrac{\pi}{2} } \;-\; i\dfrac{1}{\pi} log(A^2+B^2) \\ \because log(x+iy) \;=\; \dfrac{1}{2} log(x^2+y^2) + i tan^{-1} (\dfrac{y}{x}) \\ \; \\ \; \\ $

Comparing real and imaginary parts,

$ \therefore A \;=\; \dfrac{ tan^{-1} \dfrac{B}{A} }{ \dfrac{\pi}{2} } \\ \; \\ \; \\ \therefore \dfrac{\pi A}{2} \;=\; tan^{-1} \dfrac{B}{A} \\ \; \\ \; \\ \therefore tan( \dfrac{\pi A}{2} )= \dfrac{B}{2} \; \; \; \; \; Hence \; Proved. \\ \; \\ \; \\ Also, \; B \;=\; -\; i\dfrac{1}{\pi} log(A^2+B^2) \\ \; \\ \; \\ \therefore log(A^2+B^2) \;=\; -\pi B \\ \; \\ \; \\ A^2+B^2=e^{-\pi B} \; \; \; \; \; Hence \; Proved. $