Question

i^i^i^i^i^e=0
i^i^i^i^i^i^e=1
i^i^i^i^i^i^i^e=i
WHY

Answer 1

i^i^i = -i

Answer 2

source please.

Answer 3

actually im verifying this is true, one sec

Answer 4

I meant ((i)^i)^i... not ((i^i)^i)^i...
And I discovered this a while ago entering random queries in Wolfram Alpha, and then Google.

Answer 5

source please

Answer 6

yes i just checked, five tetrations

Answer 7

Well, I guess I can say the source is http://www.wolframalpha.com/input/?i=i%5Ei%5Ei%5Ei%5Ei%5Ei%5Ee partly.

Answer 8

What's a tetration?

Answer 9

2^2^2 is a tetration

Answer 10

a tower of powers

Answer 11

i = cos (pi/2 ) + i sin(pi/2) = e ^(i*pi/2)
therefore
i^i = [ e^(i*pi/2)]^i = e^ [ i*i*pi/2] = e^[ -pi/2]

Answer 12

then you keep doing th same thing

Answer 13

\[i^i=(e^{i\frac{ \pi }{ 2 }} )^i =e^{-\frac{ \pi }{ 2 }}\]

Answer 14

Well @perl beat me I suppose.

Answer 15

Where does that property come from, though?

Answer 16

i^i^i = [[e^(i*pi/2)]^i)^i = (e^ [ i*i*pi/2])^i = (e^[ -pi/2])^i = e^[-Pi/2*i]

Answer 17

which property?

Answer 18

\[\Large ((i)^i)^i\]etc right? You can use exponent rules to simplify some of it