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.

This would come right from Euler's formula. Let's derive it first.
There are many ways to derive it though, the Taylor series method being the most popular; here I’ll go through a different proof.
Let the polar form of the complex number be equal to $z$ .
$$\implies z = \cos x + i\sin x$$
Differentiating on both sides we get,
$$\implies \dfrac{dz}{dx} = -\sin x + i\cos x$$
$$\implies dz = (-\sin x + i\cos x)dx$$
Integrating on both sides,
$$\implies \displaystyle \int \frac{dz}{z} = i \int dx$$
$$\implies \log_e z = ix + K$$
Since $K = 0$, (Set $x = 0$ in the equation), we have,
$$\implies z = e^{ix}$$
$$\implies e^{ix} = \cos x + i\sin x$$
The most famous example of a completely real number being equal to a real raised to an imaginary is
$$\implies e^{i\pi} = -1$$
which is Euler’s identity.
To find $i$ to the power $i$ we would have to put $ x = \frac{\pi}2$ in Euler's formula.
We would get
$$e^{i\frac{\pi}2} = \cos \frac{\pi}2  + i\sin \frac{\pi}2$$
$$e^{i\frac{\pi}2} = i$$
$${(e^{i\frac{\pi}2})}^{i} = i^{i}$$
$$i^{i} = {e^{i^{2}\frac{\pi}2}} = {e^{-\frac{\pi}2}}$$
$$i^{i} = {e^{-\frac{\pi}2}} = 0.20787957635$$
This value of $i$ to the power $i$ is derived from the principal values of $\sin$ and $\cos$ which would satisfy this equation. There are infinite angles through which this can be evaluated; since $\sin$ and $\cos$ are periodic functions.
