How does $e^{i x}$ produce rotation around the imaginary unit circle?

Starting with this formulation of $e^x$ $$ e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n\tag{1} $$ and extending this definition to $e^{ix}$: $$ e^{ix}=\lim_{n\to\infty}\left(1+\frac{ix}{n}\right)^n\tag{2} $$ For a complex number $z$, let $|z|$ be its magnitude and $\arg(z)$ be its angle. If it is not already known, only a small amount of algebra and trigonometry is needed to show that $$ \begin{align} |wz|&=|w|\cdot|z|\tag{3a}\\ \arg(wz)&=\arg(w)+\arg(z)\tag{3b} \end{align} $$ Induction then shows that \begin{align} |z^n|&=|z|^n\tag{4a}\\ \arg(z^n)&=n\arg(z)\tag{4b} \end{align} Let us take a closer look at $1+\dfrac{ix}{n}$. $$ \begin{align} \left|\,1+\frac{ix}{n}\,\right|&=\sqrt{1+\frac{x^2}{n^2}}\tag{5a}\\ \tan\left(\arg\left(1+\frac{ix}{n}\right)\right)&=\frac xn\tag{5b} \end{align} $$ Using $(4a)$, $(5a)$, and $(2)$, we get $$ \begin{align} |e^{ix}| &=\lim_{n\to\infty}\left|\,1+\frac{ix}{n}\,\right|^n\\ &=\lim_{n\to\infty}\left(1+\frac{x^2}{n^2}\right)^{n/2}\\ &=\lim_{n\to\infty}\left(1+\frac{x^2}{n^2}\right)^{\frac{n^2}{2n}}\\ &=\lim_{n\to\infty}e^{\frac{x^2}{2n}}\\[12pt] &=1\tag{6} \end{align} $$ It can be shown that when $x$ is measured in radians $$ \lim_{x\to0}\frac{\tan(x)}{x}=1\tag{7} $$ Using $(4b)$, $(5b)$, and $(7)$, we get $$ \begin{align} \arg(e^{ix}) &=\lim_{n\to\infty}n\arg\left(1+\frac{ix}{n}\right)\\ &=\lim_{n\to\infty}n\arg\left(1+\frac{ix}{n}\right) \frac{\tan\left(\arg\left(1+\frac{ix}{n}\right)\right)}{\arg\left(1+\frac{ix} {n}\right)}\\ &=\lim_{n\to\infty}n\frac xn\\ &=x\tag{8} \end{align} $$ Using $(6)$ and $(8)$, we get that $e^{ix}$ has magnitude $1$ and angle $x$. Thus, converting from polar coordinates: $$ e^{ix} = \cos(x) + i\sin(x)\tag{9} $$ We get the rotational action from $(9)$ and $(3)$.

Consider a particle moving along the path $f(t)=e^{i t}$. It's instantaneous velocity is given by the derivative, and convince yourself that it is, treating $i$ as a constant, $ie^{it}$ Thus we see

$$\text{Velocity} = i\text{Position} = \text{Position (rotated by} \frac{\pi}{2} \text{radians)}$$

Because $f(0) = 1$, intitial velocity is $i$. Moving the position slightly and changing the velocity shows us that $|f(t)| = 1$ and thus $|\frac{d}{dt}f(t)|=1$. If $t =\theta$, the particle will have traveled $\theta$ radians around the unit circle.

Let's look at the $|e^{ix}|$, this is always constant $ (|\cos x + i \sin x| = \sqrt{ \cos^2x + \sin^2x} = 1)$. The only thing that is changed is $x$, now if we assign coordinates to real $(\cos x$) as $x$-coordinate and complex value $(\sin x)$ as $y$-coordinate (or imaginary axis), then this is same as parametric equation of unit circle with $x$ as parameter. As $x$ increases, the path traced by the point will be circular.

$\exp(z)$ is the function which is its own derivative. It's natural to introduce coordinates since we're thinking about the circle (a 2D figure).

We can consider it's real and imaginary parts: $\exp(iy) = c(y) + is(y)$, differentiating gives $\exp(iy) = - i c'(y) + s'(y)$ comparing with the previous gives $s'(y) = c(y)$ and $c'(y) = - s(y)$.

From the power series we find the real part is all the even powers so $c(-y) = c(y)$ and the imaginary part is all the odd powers so $s(-y) = -s(y)$, this lets us conclude Pythagoras' theorem $c(y)^2 + s(y)^2 = \exp(iy)\exp(-iy) = 1$.

From that we easily deduce that the path $(c(y),s(y))$ lies on the unit circle and is arc-length parametrized. Therefore it returns to its starting point when $y$ reaches $2 \pi$.