Powers of i

Multiply i by itself again and again. ${i}^{1} = i$ and ${i}^{2} = -1$. Then ${i}^{3} = {i}^{2}\cdot i = -i$, and ${i}^{4} = {i}^{2}\cdot {i}^{2}$, which is -1 times -1, so 1. One more factor starts over: ${i}^{5} = {i}^{4}\cdot i = i$. The powers cycle through four values forever: i, -1, -i, 1.

So for a big power, divide the exponent by 4 and keep the remainder. For ${i}^{23}$: $23 = 4\cdot 5+3$, so ${i}^{23} = {\left({i}^{4}\right)}^{5}\cdot {i}^{3}$, which is 1 times -i, or -i. A remainder of 1 gives i, 2 gives -1, 3 gives -i, and 0 gives 1.

Why 4? Because ${i}^{4} = 1$, and multiplying by 1 changes nothing, so every group of four i's drops out. That is why ${i}^{40} = 1$: 40 is 4 times 10, and no i is left over.