$\pi = 0$! It can't be! Keep in mind: I'm still in high-school so forgive my poor maths. Also remember that I'm in HIGH-SCHOOL so nothing to complex
I like to mess around with equations and I find it quite fascinating the results I can somehow come up with. Recently, I found Euler's Identity and started to mess around with this. Somehow I can up with the following result: $π = 0.$ I thought to myself, as anyone would, how is this possible?
I then decided to show my proof to my maths teacher and my parents and neither one could show me where I went wrong so I was hoping someone here could.
This is my proof:
$$e^{iπ} + 1 = 0$$
$$e^{iπ} = -1$$
$$e^{2iπ} = 1$$
$$2iπ = \ln(1) = 0$$
$$4\times-1\timesπ = 0$$
$$-4π^2 = 0$$
$$0 = 4π^2$$
$$0 = 2π$$
$$0 = π$$
Edit
As I pointed out above, my maths knowledge isn't that advanced, compared to people on this site. Therefore the answer(s) I accept and/or upvote will be based on understandability as well as how well they answer the question.
If $x$ and $y$ are real, then $e^x=e^y$ implies $x=y$. 

But if $x$ and $y$ are complex, $e^x=e^y$ only implies that
$$\frac{x-y}{2\pi i}\in\Bbb Z$$

Therefore, from $e^{2\pi i}=e^0$ we can't deduce that $2\pi i=0$.
The problem is simple. Note that if $0=\log(1)=\log(e^{2\pi I})=2\pi i$ than the logarithm is no mapping. Canonically one give a solution to this type of equations by restricting to the main branch of the complex logarithm.