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WolframAlpha seems to tell me that $e^{e^{e^{e^{e^{e^{e^{e^{e^{e^{e^i}}}}}}}}}} = 1$, see link. Is this just an error or is it for real? Adding one more $e$ to the bottom of the tower gives me the number $e$, so it's specific to the 11 $e$'s I used in the tower.

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Wolfram says it is 1.00000000000000... not just 1, so I think it is an approximation, otherwise it should have shown 1 only. – pritam Aug 23 '12 at 7:07
This happens because with 10 $e$'s you get a number very very close to zero, because with 9 $e$'s you get a number with a very large negative real part, because with 8 $e$'s you get a number whose real part is large and imaginary part is between $\pi/2$ and $3\pi/2$. The trajectory of the iterations is chaotic. – Rahul Aug 23 '12 at 7:11
up vote 27 down vote accepted

You can readily check this using an independent method. Let $x_n + i y_n\in\mathbb{C}$ be the value of a tower of $n$ copies of $e$ with a single $i$ at the top, so that $x_{n+1}+iy_{n+1}=\exp(x_n+iy_n)$. This can be rewritten as $e^{x_n}\left(\cos y_n+i \sin y_n\right)$, giving the recursion $$ x_{n+1}=e^{x_n}\cos y_n,\qquad y_{n+1}=e^{x_n}\sin y_n. $$ The starting values are $x_0=0$ and $y_0=1$. Evaluating this recursion numerically gives the following table: $$ \begin{eqnarray} (x_1,y_1) &=& (0.5403023058681398, &&0.8414709848078965) \\ (x_2,y_2) &=& (1.1438356437916404, &&1.2798830013730222) \\ (x_3,y_3) &=& (0.9002890839010574, &&3.006900083345737) \\ (x_4,y_4) &=& (-2.438030346526128, &&0.3303849520417783) \\ (x_5,y_5) &=& (0.08260952954639851, &&0.028331354522507797) \\ (x_6,y_6) &=& (1.0856817633955023, &&0.030767067267249513) \\ (x_7,y_7) &=& (2.960056578435498, &&0.09110100745978908) \\ (x_8,y_8) &=& (19.21903374615272, &&1.7557331998479278) \\ (x_9,y_9) &=& (-40856897.72613553, &&218399070.28039825) \\ (x_{10},y_{10}) &=& (-0.0, &&-0.0) \\ (x_{11},y_{11}) &=& (1.0, &&-0.0), \end{eqnarray} $$ which seems to confirm what Alpha says. However, it should be clear that what's actually happening is that a large negative real part is reached at $n=9$. This produces a numerical zero at $n=10$, followed by $1.0000\ldots$ at $n=11$. While the correct value at $n=11$ is close to $1$, it's not exact. The exact value will differ from $1$ somewhere around the $18$-millionth digit.

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Indeed, Mathematica tells me the value at $n=10$ is actually about $(-4.72 - 3.51i)\times10^{-17743926}$. – Rahul Aug 23 '12 at 7:15
Thanks, I knew it was too good to be true! I had hoped to come near of the complex fixed points $e^z = z$ but instead found this. – Dog Aug 23 '12 at 7:17
As an additional note, the tower with 10 $e$'s gives a value of $\approx -4.722810346\times 10^{-17743926}-3.513602657\times 10^{-17743926}i$; taking the exponential of that will certainly give something very nearly equal to $1$. Alternatively, substituting this tiny value into the series $\exp\,x-1=x+\frac{x^2}{2}+\cdots$ is also illustrative. – J. M. Aug 23 '12 at 7:19
Can anyone calculate this further? There seems to be an interesting pattern developing! – John Bentin Aug 23 '12 at 15:09
@JohnBentin: not exactly the same problem, but might be similar to the pattern you are refering to:… – example Sep 4 '12 at 15:05

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