As in my post about the exponential power series, I set out from the fact that the derivative of the exponential function is the exponential function itself.
Take a look at the unit circle. For any angle \(\theta\), \(\cos \theta\) is the x-value and \(\sin \theta\) the y-value of a point on this circle. Now take the derivative of both functions and stay on the same axis such that you get a new point with \(-\sin \theta\) as x-value and \(\cos \theta\) as y-value. To get to the original point, all you need is a clockwise rotation of \(\pi / 2\) around the origin.
This idea can be represented nicely in the complex plane, where the x-axis represents the real part and the y-axis the imaginary part of a complex number. To get to a point \(z\) on the circle, add the real and imaginary part \((z = \cos \theta + i \sin \theta).\) Now take the derivative of this sum. $$ \frac{d}{d \theta} \left( \cos \theta + i \sin \theta \right) = -\sin \theta + i \cos \theta $$
Multiplication with \(i\) must be a counterclockwise rotation of \(\pi / 2\) around the origin since it takes the point \((1, 0)\) to \((0, i).\) So I multiply both sides by \(1 / i\) to rotate clockwise. $$ \begin{align*} \frac{1}{i} \cdot \frac{d}{d \theta} \left( \cos \theta + i \sin \theta \right) &= \frac{-\sin \theta + i \cos \theta}{i}\\ &= \frac{-\sin \theta}{i} + \cos \theta \\ &= \cos \theta - \frac{\sin \theta (-i)}{i (-i)} \\ &= \cos \theta + i \sin \theta \end{align*} $$
Now I'm looking for another function \(f\) such that the following equation holds. $$ \frac{1}{i} \cdot \frac{d f}{d \theta} = f $$
If you know the chain rule, at this point it becomes quite clear that \(f(\theta) = e^{i \theta}\) is such a function because \(\frac{d}{d \theta} e^{i \theta} = e^{i \theta} i.\) So we have two functions that satisfy the same differential equation. $$ \begin{align*} f' &= i e^{i \theta} &= i f\\ g' &= -\sin \theta + i \cos \theta &= i g \end{align*} $$
But are they actually the same? Using the quotient rule to compute the derivative of \(f\) divided by \(g\) , we find that the rate of change of their ratio is always zero. (Dividing by \(g = \cos \theta + i \sin \theta\) is fine since it describes the points on the unit circle, so it is never zero.) $$ \begin{align*} \left( \frac{f}{g} \right)' &= \frac{f' g - f g'}{g^2}\\ &= \frac{ifg - fig}{g^2}\\ &= 0 \end{align*} $$
This means that \(\frac{f}{g}\) must be constant and we can compute this constant by evaluating the ratio at any point. Zero is easy. $$ \begin{align*} \left( \frac{f}{g} \right) (0) &= \frac{e^{i \cdot 0}}{\cos 0 + i \sin 0}\\ &= \frac{1}{1 + 0}\\ &= 1 \end{align*} $$ The ratio is \(1\) for any \(\theta\), so \(f\) and \(g\) must be equal and we end up with Euler's famous formula. $$ e^{i \theta} = \cos \theta + i \sin \theta $$
Update: I added the last section to turn this into a proper derivation.