I Am I misapplying something here? (Exponentials; Euler's identity)

[1940LaSalle]

TL;DR

Reconciling an apparent (?) paradox in an exponential

We all know from Euler's identity that exp(iπ)=-1. And from the laws of exponents, exp(2iπ)=(exp(iπ))²=1.

Further, for any real number a≠0, a⁰=1.

Then, since two things equal to the same third thing are necessarily equal, exp(2iπ)=(exp(iπ))²=a⁰=1.

Here's where I'm wondering if I've stripped one or more intellectual gears. If we now take natural logarithms, it would seem we get

ln(exp(2iπ)) = 2iπ = 0*ln (a) = 0

That would suggest at first glance that 2iπ = 0, which made me do a double take, to say the least. I realize this may well be fallacious, and I may be guilty of missing something obvious and fundamental. Help me out here (gently, if you would, please): what did I miss?

 

[PeroK]

In general, $f^{-1}(f(z))$ is not necessarily equal to $z$. Can you see why?

 

[FactChecker]

Without using complex numbers at all, you could have said that $(1)^2 = 1 = 1^0$, so $2=0$.
The truth is that the inverse of the exponential function is best understood as a multi-valued function in the complex plane. If $e^x = f(x) = 1$, then $x$ can equal $n2\pi i$, for any integer, $n$.

 

[1940LaSalle]

Thanks