i^8

i^8 = 1

Yes. Definitely there is some trick.

Remember that $i^2=-1$,

so $i^4=i^2\cdot i^2=-1\cdot -1=1$

Therefore $i^{4n}=(i^4)^n=1^n=1$, for any integer n.

$i^8=(i^4)^2=1^2=1$

$i^{25}=i^{24}\times i=(i^4)^6\times i=1\times i = i$