Imaginary paradox

Following my proof of the obvious, I thought it might be fun to prove the impossible. Math paradoxes!

Well, of course you all know about the proof of 1=0? It's included below for completeness.

Let a = b
a*a = b*a
a*a - b*b = b*a - b*b
(a+b)(a-b) = b(a-b)
a+b = b
a = 0
Let a = 1
1 = 0

The experienced algebraist quickly spots the error. One of the steps in the proof is an illegal move.

The following proof also makes an illegal move somewhere, but it breaks a rule that most people haven't ever heard of.

We start with Euler's formula:

e^ix = cos(x) + i*sin(x)

Here, "i" is used to denote the square root of -1. Euler's formula is far from a mere mathematical curiosity. I do not exaggerate when I say we use it all the time in physics. Euler's formula is the main reason that imaginary numbers are of any use at all. But I digress.

Using Euler's formula, we know...

e^iπ = -1
(e^iπ)ⁱ = (-1)ⁱ
(-1)ⁱ = e^iπi = e^-π

But then...

e^i3π = -1
(e^i3π)ⁱ = (-1)ⁱ
(-1)ⁱ = e^i3πi = e^-3π

Therefore, e^-π = e^-3π. But clearly this is wrong. Therefore, our proof is wrong. But where?