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I just learnt about the Gelfond's constant, which is $e^\pi$. Here is the part I don't understand.

We have $e^\pi=e^{-i(i\pi)}=(e^{i\pi})^{(-i)}=(-1)^{(-i)}.$ On the other hand, we have $e^\pi=e^{i\cdot(-i\pi)}=(e^{-i\pi})^{i}=(-1)^{i}.$ But obviously, we have $(-1)^i\neq (-1)^{(-i)}$.

I don't Know what went wrong. Did I made a wrong computation?

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    the law $(a^{b})^c = a^{bc}$ is quite dubious with complex numbers. But in general you did nothing too too wrong. Observe $(-1)^{-i} = ((-1)^{-1})^{i} = (-1)^i$. It's a multivalued function, but they always equal $e^{\pi + 2\pi k}$ for all $k \in \mathbb{Z}$. – 2017-02-09
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    But, $(-1)^{-i}=e^\pi$ and $(-1)^{-i}=e^{-\pi}$. They shouldn't be equal. – 2017-02-09
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    Math is full of things that shouldn't be. Basically, it all boils down to exponentiation with non-rational exponents. How do you even _define_ that? – 2017-02-09
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    @EricZhang : That's just not correct. They are both equal to both of these numbers, and many more. See my answer. – 2017-02-09

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It's not quite so obvious that $(-1)^i \neq (-1)^{-i}$; in fact, they share all of their values. What do I mean by that?

The symbols $(-1)^i$ and $(-1)^{-i}$ both represent the same set of infinitely many values:

$$(-1)^i=e^{i\log(-1)}=e^{i(\pi i + 2k\pi i)}= e^{i^2(2k+1)\pi}=e^{-(2k+1)\pi}$$ and $$(-1)^{-i}=e^{-i\log(-1)}=e^{-i(\pi i + 2n\pi i)}= e^{-i^2(2n+1)\pi}=e^{(2n+1)\pi}$$ for integral $k$ and $n$.

But these are the same sets of numbers, since the exponents are in both cases the set of odd integer multiples of $\pi$.

The error is in thinking that these expressions are single numbers.

Note: Even with real numbers, note that $(-1)^{+1}=(-1)^{-1}$, so beware of such claims of "obvious" inequality.

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    I see. Thx for pointing out that it has to do with the complex Log function. It has branch cut, so that I am looking at different branches. – 2017-02-09