Euler's formula proof with Calculus I was reading this source here and it provides a proof of Euler's formula using calculus. Although I technically understand the reasoning, I can't quite wrap my head around one particular step: if $f(x)= \cos(x)+i \sin(x)$, then $f'(x)=if(x) \implies f(x)=e^{ix}$. I can kind of get this, as derivating $e^{ix}$ gives $ie^{ix}$, and I know that $e^x$ is the only non-constant function s.t. $f'(x)=f(x)$. Is there a more clear way to think about this step? Can we extend that property of $e^x$ to when it has more in the exponent? And why does it have that property anyways? Would there be a way of obtaining this step without simply knowing that $e^{ix}$ would work?

As a complement to the explanation of @Brevan Ellefsen.
Have a look at the figure below. It explains geometrically that $e^{ix}$, imagined as a turning vector at unit speed, has a unitary "speed vector" turning at the same speed, orthogonal to it.
This illustrates the fact that differentiation is equivalent to a $\pi/2$ shift, i.e., in geometrical terms, a $\pi/2$ rotation.
If you analyse it in a separate way:
$$\begin{cases}\cos'(x)=-\sin(x)=\cos(x+\pi/2)\\ \sin'(x)=-\cos(x)=\sin(x+\pi/2)\end{cases}$$
But this boils down to say that the derivative of $e^{ix}$ is $ie^{ix}$, which is interpretated as $e^{ix}$ "turned by" $\pi/2$.

