Complex Exercise Asked by Katrina Linda about nested power of i

Exercise :

What is the value of  $p={{i}^{{{i}^{{{i}^{{{i}^{i}}}}}}}}=??$

Solution: we Know that ${{i}^{2}}=-1$

So $i={{e}^{i\frac{\pi }{2}}}\Rightarrow {{i}^{i}}={{e}^{\left( i\frac{\pi }{2} \right)i}}={{e}^{-\frac{\pi }{2}}}$

$w={{i}^{i}}={{e}^{-\frac{\pi }{2}}}\Leftrightarrow m={{i}^{{{i}^{i}}}}={{i}^{w}}={{i}^{{{e}^{\frac{-\pi }{2}}}}}={{\left( {{e}^{i\frac{\pi }{2}}} \right)}^{{{e}^{-\frac{\pi }{2}}}}}={{e}^{i\frac{\pi }{2}{{e}^{\frac{-\pi }{2}}}}}={{e}^{iw\frac{\pi }{2}}}$  (1)

So $n={{i}^{{{i}^{{{i}^{i}}}}}}={{i}^{m}}={{\left( {{e}^{i\frac{\pi }{2}}} \right)}^{m}}$

Thus $p={{i}^{{{i}^{{{i}^{{{i}^{i}}}}}}}}={{i}^{n}}={{\left( {{e}^{i\frac{\pi }{2}}} \right)}^{n}}={{e}^{in\frac{\pi }{2}}}$

So $in=i{{\left( {{e}^{i\frac{\pi }{2}}} \right)}^{m}}={{e}^{i\frac{\pi }{2}}}\times {{e}^{im\frac{\pi }{2}}}={{e}^{i\frac{\pi }{2}+im\frac{\pi }{2}}}={{e}^{i\frac{\pi }{2}\left( 1+m \right)}}$   (2)

i.e $in\frac{\pi }{2}=\frac{\pi }{2}{{e}^{i\frac{\pi }{2}\left( 1+m \right)}}$

Hence $iw={{e}^{i\frac{\pi }{2}}}{{e}^{-\frac{\pi }{2}}}={{e}^{\frac{\pi }{2}\left( i-1 \right)}}$ thus $iw\frac{\pi }{2}=\frac{\pi }{2}{{e}^{\frac{\pi }{2}\left( i-1 \right)}}$        (3)

Therefore , $p={{e}^{in\frac{\pi }{2}}}={{e}^{\frac{\pi }{2}{{e}^{i\frac{\pi }{2}\left( 1+m \right)}}}}={{e}^{\frac{\pi }{2}{{e}^{i\frac{\pi }{2}\left( 1+{{e}^{\frac{\pi }{2}{{e}^{\frac{\pi }{2}\left( i-1 \right)}}}} \right)}}}}$$p=0.387166181086115+0.030527081605484428i$

${{i}^{{{i}^{{{i}^{{{i}^{i}}}}}}}}=\exp \left[ Pi/2*\exp \left[ \left( I*Pi/2 \right)\left( \left( 1+\exp \left[ Pi/2*\exp \left[ Pi/2\left( I-1 \right) \right] \right] \right) \right) \right] \right]$

By using Mathematica 10.0 we get the following approximation       Q.E.D