Here we are examining the imaginary number ‘i’, defined as the square root (sqrt) of ‘-1’.We begin by using Euler’s identity:e^(iπ)+1=0e^(iπ)=-1Since sqrt(x) is the same as x^(1/2):(e^(iπ))^(1/2)=sqrt(-1)Using (a^b)^c=a^(bc) and the definition of i=sqrt(-1):e^(iπ/2)=iThen to achieve i^i as specified in the question:(e^(iπ/2))^i=i^iUsing (a^b)^c=a^(b*c) again:e^(-π/2)=i^iSince the LHS has no imaginary part, the RHS is also real. We have proven that i^i is real and equal to e^(-π/2).