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.

Write $i=e^{\frac{\pi}{2}i}$, then $i^i=(e^{\frac{\pi}{2}i})^i = e^{-\frac{\pi}{2}} \in \mathbb{R}$. Be careful though, taking complex powers is more... complex... than it may appear on first sight $-$ see here for more info.
In particular, it's not well-defined (until we make some choice that makes it well-defined); we could just have well written $i=e^{\frac{5\pi}{2}i}$ and obtained $i^i=e^{-\frac{5\pi}{2}}$. But $i^i$ can't be equal to both $e^{-\frac{\pi}{2}}$ and $e^{-\frac{5\pi}{2}}$ can it?
Despite the lack-of-well-defined-ness, though, $i^i$ is always real, no matter which '$i^{\text{th}}$ power of $i$' we decide to take.

More depth: If $z,\alpha \in \mathbb{C}$ then we can define
$$z^{\alpha} = \exp(\alpha \log z)$$
where $\exp w$ is defined in some independent manner, e.g. by its power series. The complex logarithm is defined by
$$\log z = \log \left| z \right| + i\arg z$$
and therefore depends on our choice of range of argument. If we fix a range of argument, though, then $z^{\alpha}$ becomes well-defined.
Now, here, $z=i$ and so $\log i = i\arg i$, so
$$i^i = \exp (i \cdot i\arg i) = \exp (-\arg i)$$
so no matter what we choose for our range of argument, we always have $i^i \in \mathbb{R}$.
Fun stuff, eh?
