TITLE: Interpreting the Famous Five equation
QUESTION [32 upvotes]: $$e^{\pi i} + 1 = 0$$
I have been searching for a convincing interpretation of this. I understand how it comes about but what is it that it is telling us? 
Best that I can figure out is that it just emphasizes that the various definitions mathematicians have provided for non-intuitive operations (complex exponentiation, concept of radians etc.) have been particularly inspired. Is that all that is behind the slickness of the Famous Five equation?
Any pointers?

REPLY [3 votes]: A way to introduce the exponential is by the ODE:
$$f' = \alpha f, \ \ f(0) = 1,$$
whose unique solution is $t \mapsto exp(\alpha t)$. This is common when $\alpha$ is natural, but what if $\alpha$ is complex? And, in particular, what if $\alpha = i$? Of course, $f$ must be complex-valued instead of real-valued.
Then $f' = if$, that is, $f'$ is the vector $f$ rotated by $\pi/2$ in the trigonometric direction. If we forget a moment about the complex structure, we get:
$$y' = x,  \ x'=-y, \ \ f(0) = (x,y)(0) = (1,0).$$
A solution of this system is a trajectory whose speed is always tangential to its position vector. But then, the function $(x,y) \mapsto x^2+y^2$ is a first integral, so it is constant along trajectories. In particular, the trajectory starting from $(1,0)$ stays on the circle of radius $1$ centered on $0$. But then, since $|f|$ is constant, so is $|f'|$: the trajectory moves at unit speed along this circle of radius $1$.
Since the trajectory starts from $1$, after a time $\pi$, it will have traveled along a half-perimeter, so it will be in $(-1,0)$. Whence $exp(i \pi) = -1$.
@Andrey Rekalo's interpretation of this equation is Euler's method applied to this ODE.