complex numbers - Error in proof that $1 = -1$

I have created a proof that$1 = -1$ but I know that this is impossible. Could someone help me find the flaw in this proof...

$i = \sqrt{-1}$

Given

$i^2 = -1$

Given

$i^4 = 1$

Given
$i^8 = 1$

Given
--------------------------All Common Knowledge Above

$i^4= i^8$

Take sqrt of both sides...

$i^2=i^4$

Take sqrt of both sides...

$i= i^2$

$i=-1$

$i^2= -1$ (sub for $i$)

$-1 \times -1$ (sub for $i^2$) = $1$

$1=-1$

Thankyou all for helping me. I looked at the other questions and this question is not a duplicate. However, we all have one common error; we forgot +- when taking the square root of i^4 = i^2

Answer

What people seem to be calling you out for without explanation is that if $a^2 = b^2$, then we can have that $a = \pm b$. We can't know which of $b$ or $-b$ we started with, though.

So when you say that $i^4 = i^8$, then good. You're on the right track. But your next step needs to be that $i^2 = \pm i^4$. You then have no contradiction because one of $i^4$ and $-i^4$ is certainly equal to $i^2$.

I hope this helps you. Always try things like these, even if others call you out for being silly. Before being amazing you have to be a little silly.