Prove that $i^i$ is a real number According to WolframAlpha, $i^i=e^{-\pi/2}$ but I don't know how I can prove it.
Here's a proof that I absolutely do not believe: take its complex conjugate, which is $\bigl({\bar i}\bigr)^{\bar i}=(1/i)^{-i}=i^i$. Since complex conjugation leaves it fixed, it’s real!

EDIT: In answer to @Isaac’s comment, I think that to justify the formula above, you have to go through exactly the same arguments that most of the other answerers did. For complex numbers $u$ and $v$, we define $u^v=\exp(v\log u)$. Now, the exponential and the logarithm are defined by series with all real coefficients; alternatively you can say that they are analytic, sending reals to reals. Thus $\overline{\exp u}=\exp(\bar u)$ and $\overline{\log(u)}=\log\bar u$. The result follows, always sweeping under the rug the fact that the logarithm is not well defined.
Let say: $f(z) = z^\theta $
We know Euler's formula: $e^{i \theta} = \cos(\theta) + i \sin(\theta)$
Using it we will get: 
$$z^\theta = e^{\theta \ln(z)} = e^{\theta (\ln\|z\| + i\arg (z))} = e^{\theta \ln\|z\| }e^{ i \theta \arg (z))}= e^{\theta \ln\|z\| }{(\cos(\theta\arg (z)) + i \sin(\theta\arg (z)))} z \in C$$ 

So if $$z = i \wedge \theta = i \implies$$ $$ z^\theta = i^i = e^{i \ln(i)} = e^{i (\ln\|i\| + i\arg (i))} = e^{i \ln\|i\| }e^{ i i \arg (i))}= e^{i \ln(1) }e^{- \frac{\pi}{2}+2\pi k}= e^0 e^{- \frac{\pi}{2}+2\pi k}=e^{- \frac{\pi}{2}+2\pi k} $$ which is a bunch of Real numbers depending $k \in \mathbb Z$  

So it is already proved that $i^i$ is a real number.