How is $ i^{-1} = -i$ and $i^{-3} = i$? Now I know that with positive powers of $i$ the cycle is:  $i , -1 , -i , 1\ldots$
The negative power cycle is: $-i , -1 ,  i , 1 \ldots$
Can someone explain to me how $\frac 1 {\sqrt{-1}}$ is equal to $-i$ and $\frac 1 {-\sqrt{-1}}$ is equal to $i$?

$\frac{1}i=i^{-1}$ is true by definition and $n+\frac{1}n=\frac{n^2+1}n$ (since $n=\frac{n^2}n$ and $\frac{n^2}n+\frac{1}n=\frac{n^2+1}n$) is also true. If you plug $i$ in $\frac{i^2+1}i=\frac{-1+1}i=\frac{0}i=0$.
Therefore $i+\frac{1}i=0$ and $\frac{1}i=-i$. By our original statement (namely $i^{-1}=\frac{1}i$) we can conclude that $i^{-1}=-i$. 
$i^{-3}=\frac{1}{i^{3}}$ by definition also and $n- \frac{1}{n^3}=\frac{n^4-1}{n^3}$ (since $\frac{n^4}{n^3}-\frac{1}{n^3}$ clearly is $\frac{n^4-1}{n^3}$) when we plug $i$ into this $\frac{i^4-1}i = \frac{1-1}{i} = \frac{0}i = 0$. 
Thus, $i - \frac{1}{i^3}$ = 0. This becomes $\frac{1}{i^3}=i$. Therefore $i^{-3} = i$.
