$i^{i^{i^{i...}}} = A+iB$ .....(i) Taking loge on both sides, $i^{i^{i^{i...}}} logi=log(A_+iB)$ $(A+iB)logi=log(A+iB)$ But we know that,in polar form, $i=e^{ipi/2}$           therefore, $(A+iB)log(e^{ipi/2})=log(A+iB)$ therefore,$(A+iB)ipi/2=log(A+iB)$ ${Aipi\over2}-{Bpi\over2}=log(A+iB)$ Raising both sides to power of e, we get, $e^{{-Bpi\over2}+{Aipi\over2}}=A+iB$ On comparing with standard polar form $a+ib=re^{itetha}$ we get,$\sqrt{{A^2+B^2}}=e^{-ipi\over2}$ therefore, $A^2+B^2=e^{-piB}$ And angle $titha=tan^{-1}(B/A) $ therefore, $tan^{-1}(B/A)={Api\over2}$ therefore. $tan(Api/2)={B/A}$

HENCE PROVED!