We wish to prove that
$$e^{i\pi}+1=0.$$


The standard approach is to use Euler's formula (immediate, for example, from the series definition of the exponential, sine and cosine) and then to use the *facts* that,
$$\sin(\pi)=0\text{, and }\cos(\pi)=-1$$
to draw the necessary conclusion.


*But*, starting with the series definitions of sine and cosine, I struggle to conclude that
$$\sin(\pi)=0\text{, and }\cos(\pi)=-1.$$


The goal is to show that
$$\sum\_{n=0}^\infty\frac{(2\pi i)^{2n+1}}{(2n+1)!}=0,$$
and that
$$\sum\_{n=0}^\infty\frac{(2\pi i)^{2n}}{(2n)!}=-1.$$


Attempts by me to prove these relations always end up back where I started. My feeling is that it must somehow come down to some fundamental geometric consideration of the polar description of $\mathbb{C}$ - after all, what is $\pi$ but the circumference of a circle divided by its diameter? Somehow this must be mentioned in any valid proof.


I feel that this issue is not always taken into consideration when a *proof* of Euler's identity is presented.


Something to gawk at; a Mathematica notebook which plots the partial sums of $e^{ix}$ in the complex plane: [ExponentialPartialSumsPlotted.nb](http://www.ms.unimelb.edu.au/~smcateer/ExponentialPartialSumsPlotted.nb)



Rigourous proof of Euler's identity.
Denote by $f(x)$ the point on the unit circle with angle $x$. A simple geometric consideration shows that $f'(x)=if(x)$ which expresses the fact that any tangent line to the unit circle is perpendicular to the corresponding radius. Now $f^{(n)}(x)=i^nf(x)$ by induction, hence $f^{(n)}(0)=i^n$, and Taylor's formula with remainder term shows that $f(x)=\sum\_{n=0}^\infty (ix)^n/n!$. This means that if we define the complex exponential function with the usual Taylor series, then $f(x)=e^{ix}$. Now $e^{i\pi}=-1$ is immediate, once we define $\pi$ as the half-perimeter of the unit circle.


