Why is $e^{2πi} = e^0$ true while $2πi = 0$ is false? I came across a perplexing thing while testing some assumptions in Wolfram|Alpha, and as I don't have a math education beyond college algebra, I thought this would be a good place to ask. I would just like to emphasize that this is not a homework question, and is purely curiosity.
Common faulty math proofs tend to use logic like this:
1^0 = 1
2^0 = 1
therefore 1 = 2

This is obviously false because $x^0$ is defined as 1 for all real values of $x$.
But in the case of $e^{2\pi i} = e^0$, the base is the same: $e$. So if $a^x = a^y$ is true, shouldn't it follow, then, that $x = y$ and therefore $2\pi i = 0$? If not, why?
 A: That's because the exponential isn't an injective function on $\mathbb C$. If is an injective function on $\mathbb R$, and so the single-valued logarithm is well-defined on $\mathbb R^+$ but not on $\mathbb C$.
This isn't true for the same reason that $(-1)^2=1^2$ doesn't imply that $-1=1$.
A: Recall that
$$f(x)=f(y) \implies x=y$$
$\forall x,y$ in the domain requires that $f$ is an injective function.
A: Simply because the exponential function is periodic with period $2\pi i$.
That means that for any $x$, the values of $e^x$ and $e^{x+2\pi i k}$ are equal for all integers $k$.
In particular, you are looking at $x=0$ and $k=0,1$; so $e^0=e^{2\pi i}$, but of course $0$ and $2\pi i$ are different numbers.
A: I think the others have given fine answers, but since they haven't been accepted I'll offer one more.  Perhaps the piece you're missing (I'm guessing) is this:

So if $a^x=a^y$ is true, shouldn't it follow, then, that $x=y$?

No, it doesn't follow!  Suppose e.g. that $a=1$: clearly, $1^5 = 1^2$ does not imply $5 = 2$ (the former is true and the latter is false).  So one cannot simply equate the exponents (unless one knows that some additional constraints hold).
What then is the mechanical ("algebraic") procedure for resolving expressions like $a^x=a^y$?  Take $\log_a$ of both sides, i.e. the logarithm with base $a$.  So for $e^x = e^y$, you would apply $\log_e = \ln$, the natural logarithm.  And now here you apply what the others have said: just like taking a square root (to solve $x^2 = y^2$, yielding $y = \pm x$), or applying an inverse sine (to solve $\sin(x) = \sin(y)$, yielding $x = y \pm 2n\pi$), when applying the logarithm you have to account for the fact that the $e^{\text{a complex number}}$ is not one-to-one (multiple inputs can mapped to the same output), unlike $e^{\text{a real number}}$.  In particular, you get $x = y \pm i2n\pi$.  Is this consistent with $e^0 = e^{i2\pi}$?  Phew, yes.
