# Simplifying Long Imaginary Number Expression

I came across this problem that I couldn't really find an effective approach for. It was the simplification of $(i+1)^{2010} \text{ }- (i-1)^{2010}.$

I know that this roughly translates to $(i^{2010} + i^{2019} + ... i + 1) \text{ }$ minus a similar expression, and I also know that the answer is $2^{1006}i$, but I just don't know any other steps that I could take.

$$(i\pm1)^2=\pm2i$$
$$(i\pm1)^{4n+2}=(\pm2i)^{2n+1}=\pm2i((\pm2i)^2)^n=\pm2i(-4)^n$$