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

reallya proof - it's a definition. Maybe it should be Euler's definition of \pi; Let \pi be the real number such that e^(i\pi)+1=0. To call something an identity for my part would require that \pi has been used as an input (coming from some separate definition) somehow. – Stephen Mc Ateer Jan 6 '11 at 21:57derivethat $2\pi r$ is the circumference of the circle radius $r$. This is simply doing things in the other direction. – Eric Naslund Feb 25 '11 at 4:40