Why does rearranging Euler's identity in this manner result in a false statement?

Question

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?

Answer 1

You have $$ \sin(2\pi)=\sin(0). $$ Then $2\pi=0$?

The logarithm, if you care to define it, is not one-to-one.