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 $i$?
 A: For the first part of your question, multiply the fraction by $\frac{i}{i}$ & observe that:
$$\frac{1}{i}=\frac{i}{i\cdot i}=\frac{i}{-1}=-i$$
For the second part of the question, observe that:
$$\frac{1}{-i}=-\frac{1}{i}$$
Then use the reasoning that that $\frac{1}{i}=-i$ from the part one. If I've misunderstood what you're asking, please feel free to tell me (with more brackets, so I can see what you want).
I've edited all traces of $\sqrt{-1}$ to $i$ in response to Robjohn's comments. In future, it would be helpful if other users are less disrespectful when disagreeing with each other. Thanks.
A: $$i^{-1}=\frac1i=\frac i{i\cdot i}=\frac i{i^2}=\frac i{-1}=-i\quad\text{QED}$$
$$i^{-3}=\frac1{i^3}=\frac{i}{i^3\cdot i}=\frac{i}{i^4}=\frac{i}1=i\quad\text{QED}$$
A: $\frac{1}i=i^{-1}$ 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 $i^{-3} = i$.
A: You know that $i^2 = -1$ (this is almost the definition of $i$, and anyway you've already written as much with your list of powers of $i$), and rearranging gives
$i(-i) = 1$,
so by definition $i^{-1} = -i$. Similarly, factoring the left-hand side of $i^4 = 1$ gives
$i(i^3) = 1$,
so
$i^{-3} = (i^3)^{-1} = i$.
Though it's not a proof, exactly, there's a useful way to think about multiplication by (powers of) $i$, and indeed, complex numbers in general. Multiplying any complex number $a + bi$ (which we can visualize as the point $(a, b)$ in the Cartesian plane) by $i$ gives
$i(a + bi) = ai + b i^2 = -b + ia$,
which we can visualize as the point $(-b, a)$. But this is exactly the point we get when we rotate our starting point $(a, b)$ anticlockwise about the origin a quarter-turn! (To see this, just note that $(a, b)$ and $(-b, a)$ are the same distance from the original, and the line segments connecting them are perpendicular; we can see the latter using the dot product or the fact that the slopes of the segments have product $-1$, provided our first point wasn't on a coordinate axis.)
So, multiplying by $i^2$ is rotates a point a quarter turn twice, for a total of a half-turn about the origin (or just reflects it through the origin if you like), multiplying by $i^3$ rotates a point three-quarters of a turn around the origin, and multiplying by $i^4$ rotates a point a full turn around the origin, but this is the same as not rotating the point at all. Since $i(-i) = 1$, multiplying by $-i = i^{-1}$ undoes the transformation given by multiplying by $i$, and so it must rotate points clockwise about the origin. With these facts in hand, both your useful sequences of powers of $i$ and the two identities you asked about become intuitive statements about rotating some multiple of quarter-turns.
If this sort of thing interests you, you might like to try to work out the geometric meaning of multiplying by a general complex number $c+di$. This generalizes further to quarternions, a number system that includes the complex numbers, as well as new directions parametrized by new constants $j$ and $k$, and which can be used to track rotations in three-dimensional space.
