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 -
$$-i^3=\left(\frac1i\right)^3=\frac{1}{i^3}=\frac{1}{-i}=-(-i)=i$$

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.
