Why is $\left(e^{2\pi i}\right)^i \neq e^{-2 \pi}$?

Here’s my (obviously flawed) proof that $1=e^{-2 \pi}$:
e^{2 \pi i} = 1\\
\left(e^{2\pi i}\right)^i = 1^i\\
e^{-2 \pi} = 1

What’s the issue? I understand that exponentiation is not injective (and thus $-1 \neq 1$ even though $(-1)^2 = 1^2$), but I don’t think that’s an issue here: I’m only raising things to the power of $i$, which I don’t think is multi-valued.

Solutions Collecting From Web of "Why is $\left(e^{2\pi i}\right)^i \neq e^{-2 \pi}$?"

In complex numbers, either $x^y$ is a multivalued function, or you have to give up the notion that $(x^y)^z = x^{yz}$.

If you allow $x^y$ to be multi-valued, then one of the values for $1^i$ is $e^{-2\pi}$.

If $x^y$ is not multivalued, then you have to pick a single value for $\log 1$ to define $1^y$. We usually pick $\log 1 = 0$, for some reason. 🙂

The multivalued nature makes sense when you consider that $\sqrt{1}=1^{1/2}$ can be thought of as having two values, $-1$ and $1$. In general, though, when $y$ is irrational, you get $1^y$ (or more generally, $x^y$) can take infinitely many values.

The only time $x^y$ is naturally single-valued is when $y$ is an integer.

The law of exponents $x^{ab} = (x^a)^b$ does not hold in general.

Raising numbers to complex powers is multivalued.

For example, $1^i = e^{\log(1^i)} = e^{i (\ln(1)+2\pi k i)} = e^{i(2\pi k i)} = e^{-2\pi k}$ for any integer $k$. Choosing $k=-1$ matches your left hand side, $k=0$ matches the right.