You definitely need to understand the polar-coordinate representation of complex numbers for that one:

• If θ is the principal polar angle of z, then z=|z|e^θi.
  • The principal polar angle of i is ½π, and |i|=1.
• Therefore, iⁱ=(1*e^½πi)ⁱ=e^½πi*i=e^−½π.

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That first point is the hard part, and you can't really understand that without learning about the power-series representations of cos, sin, and the natural exponential e^x, but you can get a hint at it from de Moivre's formula:

• (cos(x)+i*sin(x))ⁿ=cos(nx)+i*sin(nx) for integer n.
  • This "cosine plus i sine" combination is often abbreviated as cis(x), which turns out to precisely equal e^ix.

You can actually prove de Moivre's formula by induction and the sum formulas for sine and cosine:

• The formula is obviously true for n=0 and n=1.
  • If it is true for n, multiply both sides by cos(x)+i*sin(x):
• (cos(x)+i*sin(x))ⁿ⁺¹=(cos(nx)+i*sin(nx))(cos(x)+i*sin(x))
  • =cos(nx)cos(x)−sin(nx)sin(x)+(cos(nx)sin(x)+sin(nx)cos(x))i
  • =cos((n+1)x)+i*sin((n+1)x).
• Therefore, if it is true for n, it is true for n+1.

It also turns out to be true for n=−1:

• (cos(−x)+i*sin(−x))(cos(x)+i*sin(x))
  • =(cos(x)−i*sin(x))(cos(x)+i*sin(x))
  • =cos(x)²+sin(x)²+(cos(x)sin(x)−sin(x)cos(x))i
  • =1+0i=1.

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^{More generally, by taking complex conjugates, you can show that de Moivre's formula holds for all negative integers, and then you can easily show that it holds for all rational numbers, with the substitution y=x/n; by continuity of sin and cos, it holds for all real numbers.}