What does the square root of minus $i$ equal? Can you enter the rabbit hole recursively?
If the $ \sqrt{-1} = i $ then, what does $ \sqrt{-i} $ equal?

Let us assume there exist a complex number $m$ such that $m * m = -i$, then $m$ will be the answer we are looking for $\sqrt{-i}$.
we can expand $m$ as $a + i.b$, then we have $(a + i.b)^2 = -i$; expanding we get
$a^2 - b^2 + i.2ab = -i$   $[i^2 = - 1] $; to make this true we must have
$a^2 - b^2 = 0$ --> (1)
and 
$a * b = -1/2$ --> (2)
From (2) we have $b = -1/2a$ substituting in (1) we have
$a^2 - (-1/2a)^2 = 0$
rearranging we get the below
$a2 - (1/4a^2) = 0$
Multiplying both sides of equation with $a^2$ we get
$a^4 = 1/4$ and so $a=\pm1/\sqrt{2}$ we know $b=-1/2a$
So, when $a = 1/\sqrt{2}$ then $b = - 1/\sqrt{2}$ 
like wise when $a = -1/\sqrt{2}$ then $b = 1/\sqrt{2}$
Finally substituting the $a$ and $b$ values we get two roots for $\sqrt{-i}$
$$(-1/{\sqrt{2}}) + i.(1/\sqrt{2})$$ and $$(1/\sqrt{2}) - i.(1/\sqrt{2})$$ 
You'll see squaring any of the above solution you'll get $-i$
