TITLE: Why does rearranging Euler's identity in this manner result in a false statement?
QUESTION [5 upvotes]: I placed each of the following steps into Wolfram alpha after working it out in my head. All steps prior to the one marked with the (*) held true.
$$e^{i \pi} = -1$$
$$e^{2i \pi} = 1$$
$$\ln(e^{2i \pi}) = \ln(1)$$
$$\tag {*} 2i\pi = 0$$
Every conclusion that can be drawn from the last statement is trivially false (such as $i = 0$ or $1 = 0$ etc.). Where did I go wrong?

REPLY [6 votes]: You have $$
\sin(2\pi)=\sin(0).
$$
Then $2\pi=0$? 
The logarithm, if you care to define it, is not one-to-one.