Question 133841
{{{red(i^0)=1}}}  because anything to the zero power is 1
{{{red(i^1)=i}}}  because anything to the 1 power is itself
{{{red(i^2)=-1}}}  by definition of i
{{{red(i^3)=i^2*i^1=-1*i=-i}}}  because {{{a^n*a^m=a^(n+m)}}}
{{{i^4=i^2*i^2=-1*-1=1=red(i^0)}}}
{{{i^5=i^4*i^1=1*i=i=red(i^1)}}}
{{{i^6=i^4*i^2=1*-1=-1=red(i^2)}}}, and 
{{{i^7=i^4*i^3=1*(-i)=(-i)=red(i^3)}}}


And the pattern repeats forever.


So, the process is to take the exponent on i, 55 in this case and perform integer division with a divisor of 4.  Integer division is the first kind of division you learned in the 2nd or 3rd grade where you got a quotient and a remainder.  Then raise i to the power of the remainder.


{{{55/4=13r3}}}, so {{{i^55=i^3=-i}}}