Is ${(-1)}^{\frac{1}{6}}=i$? I just thought of this and tried to simplify. ${(-1)}^\frac{1}{6}$.
$${(-1)}^\frac{1}{6}={[{(-1)}^\frac{1}{3}]}^{\frac{1}{2}}$$
Since the cube root of $-1$ is $-1$,
$${(-1)}^{\frac{1}{6}}=\sqrt{-1}\to{(-1)}^{\frac{1}{6}}=i$$
Is this a correct conclusion? If so, does this work to every roots with even index, where the radicand is negative?
 A: If $z = (-1)^\frac 16$ then $z^6 = -1$ or $z^{6} + 1 = 0$
This we can factor!
$(z^{2} + 1)(z^4 - z^2 + 1) = 0$
From the left hand factor, $z = \pm i$ are indeed solutions.  
The right hand factor is a little trickier.
$z^4 - z^2 + 1 = (z^2 + \sqrt 3 z + 1)(z^2 - \sqrt 3z + 1)$
$z = \pm\frac {\sqrt {3}}{2} \pm \frac {1}{2} i$
There are 6 solutions.
These are called the roots of unity.
There are other methods to find these roots.
If you know De Moivre's law you can apply it.
$z^n = |z|^n(\cos \theta + i\sin \theta)^n = |z|^n(\cos n\theta + i\sin n\theta)$
Using this approach:
$(-1)^\frac 16 = (\cos \pi + i\sin \pi)^\frac 16 = \cos \frac {\pi}{6} + i\sin \frac {\pi}6$
$\cos \frac {(2k-1)\pi}{6} + i\sin \frac {(2k-1)\pi}6$ are also roots.
And by Euler formula
$e^{\pi i} = -1\\
e^{\frac {\pi}{6} i} = (-1)^\frac 16$
A: It is true that $i^6=-1$, but fractional powers of negative numbers are not uniquely defined, 
and the "general rule" $(a^m)^n=a^{m×n}$ does not always work when $m$ and $n$ are not integers.  
There are six complex $6^{th}$ roots of $-1$; they are  $\pm i$ and $\pm\dfrac{\sqrt3} 2\pm \dfrac i2$.
