Very funny indeed!

Let $x$ be the $i^{th}$ root of $i$

$x^i=i$

Now, $i=e^{i\frac {π}{2}}$

Thus, $x=e^{\frac {π}{2}}$

Relevant wiki: Euler's Formula

By Euler's formula: $e^{\theta i} = \cos \theta + i\sin \theta$ , $\implies i = e^{\left(2k + \frac \pi 2\right)i}$ , where $k$ is an integer. Therefore, $i^\frac 1i = e^{\left(2k+\frac \pi 2\right)i \times \frac 1i} = e^{2k+\frac \pi 2}$ .

And the solution available is $\boxed{e^\frac \pi 2}$ .

Hint: Use e^pi*i=-1