Why is $-i^3 = i$? Why is the value of $-i^3$ equal to $i$?
After experimenting, I got this result -
$-i^3=-i^2\cdot -i=1 \cdot -i=-i$
What is the error in my proof?
EDIT
Here is the original proof -
$$-i^3=\left(\frac1i\right)^3=\frac{1}{i^3}=\frac{1}{-i}=-(-i)=i$$
 A: $$i^3=i^2 \cdot i=(-1)\cdot i=-i$$ so that $$-i^3=-(-i)=i$$
You are going wrong by saying that $-i^3=-i^2\cdot (-i)$. 
A: Wondering if you meant $(-i)^3$ or $-(i^3)$?
For the former, $(-i)^3=-i \times -i \times -i=-(i^3)=-[(i^2)(i)]=-[(-1)(i)]=i$. Note that $i^2=-1$.
For the latter, $-(i^3)=-[(i^2)(i)]=-[(-1)(i)]=i$.
So, both are the same; i.e., $(-i)^3=i=-(i^3)$.
A: Note that
$$
-i^3=-(i^2\cdot i)=-(-1\cdot i)=-(-i)=i
$$
You're computing $(-i)^3$ because you're interpreting wrongly the symbol $-i^3$. In the standard notation, exponentiation has precedence over taking the negative, so
$$
-a^n
$$
should be interpreted as “the negative of ($a$ raised to the power $n$)”, not “(the negative of $a$) raised to the power $n$)”.
The error is clearly shown when you write
$$
-i^3=\left(\frac{1}{i}\right)^{\!3}
$$
which is wrong for the reason above.
However, your error is subtler (as noticed by Darth Geek in comments): you're not consistent with your interpretation when you write
$$
-i^3=-i^2\cdot -i
$$
and then substitute $-i^2$ with $1$. But the main problem is that you're assuming $-a^n=(-a)^n$, which is wrong: consider $-1^n$ and you'll understand why. $-1^n=-1$, whereas $(-1)^n$ is $1$ when $n$ is even and $-1$ when $n$ is odd.
A: $-i^3=-i^2 . i=-(-1).i=i$
( Note that $ i^2=-1$)
A: Looking at the complex argument of $-i^3$ we see the following things: 
$$-i^3=(-i)^3=(e^{-\frac{\pi}{2}i})^3=e^{3\left(-\frac{\pi}{2}\right)i}=$$
$$e^{-\frac{3\pi}{2}i}=e^{\left(2\pi+\left(-\frac{3\pi}{2}\right)\right)i}=e^{\frac{\pi}{2}i}=i$$
