Prove that $e^{i\pi} = -1$

When I first found out that $e^{i\pi} = -1$, I was blown away. Does anyone here know one of (many I'm sure) proofs of this phenomenal equation? I can perform all of the algebra to get the $-1$. But, where does this come from? What is the derivation?

Here's a slick derivation. Let $f(x)= e^{-ix}(\cos x + i\sin x)$. Taking the derivative we have that $f'(x) = -ie^{-ix}(\cos x + i \sin x) + e^{-ix}(-\sin x + i\cos x) = 0$ so $f(x)$ is constant. But $f(0) = e^0(1 + 0) = 1$ so $f \equiv 1$ so $e^{ix} = \cos x + i \sin x$. Plugging in $x = \pi$ yields the result.

Remark: This proof is very beautiful, and uses the characteristic property of everything involved ($e^x$ is the unique nontrivial function that is its own derivative, $i$ squares to $-1$, etc), but as a first-timer the Taylor series proof told me much more.

The identity is a special case of Euler's formula from complex analysis, which states that

$e^{ix}$=$\cos x+i\cdot\sin x$

for any real number $x$. (Note that the variables of the trigonometric functions sine and cosine are taken to be in radians, and not in degrees.) In particular, with $x = \pi$, or one half turn around the circle:

$e^{i\pi}$=$\cos\pi+i\cdot\sin\pi$

Since

$\cos\pi=-1$

and

$\sin\pi=0$

it follows that

$e^{i\pi}$ =$-1+0i$

which gives the identity

$e^{i\pi}$$+1=0$

The physical explanation of Euler's identity is that it can be viewed as the group-theoretical definition of the number $\pi$. The following discussion is at the physical level but can be made mathematically strict. The group is the group of rotations of a plane around 0. In fact, one can write:

$e^{i\pi}$=$(e^{i\delta})^{\pi/\delta}$

with $\delta$ being some small angle. The last equation can be seen as the action of consecutive small shifts along the circle caused by the application of infinitesimal rotations starting at 1 and going for the total length of the arc connecting points 1 and -1 in the complex plane. In fact, each small shift can be written as multiplication by

$1+i\delta$

and the total number of shifts is π/δ. In order to get from 1 to -1 the total transformation would be

$(1+i\delta)^{\pi/\delta}$

Now, taking the limit when $\delta \to0$, denoting $i\delta = 1/n$ and using the definition of:

$e$=$\lim\limits_{n \to \infty}(1+\frac{1}{n})^n$

we arrive at Euler's identity. The $\pi$ itself is defined as the total angle which connects $1$ to $-1$ along the arch.

Summarizing, we can say that because the circle can be defined through the action of the group of shifts which preserve the distance between a point and another point, the relation between π and e arises. This simple argument is the key to understanding other seemingly miraculous relations involving $π$ and $e$.

Source: Wikipedia

$$e^{ix}=\sum_{n=0}^{\infty}\frac{(ix)^n}{n!}=\sum_{k=0}^{\infty}\frac{(ix)^{2k}}{(2k)!}+\sum_{k=0}^{\infty}\frac{(ix)^{2k+1}}{(2k+1)!}=$$ $$=\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k}}{(2k)!}+i\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k+1}}{(2k+1)!}=$$ $$=\cos x+i\sin x$$ since $$i^{2k}=(-1)^k,i^{2k+1}=i(-1)^k$$ for $x=\pi$ we get $$e^{i\pi}=\cos (\pi)+i\sin(\pi)=-1$$