How do complex number exponents actually work?
This is more an answer to the title of your question. A possible (and perhaps the standard) way of properly defining things is as follows. See any reasonable rigorous textbook on analysis for the proofs.
First define the exponential function $\exp : \mathbb{C} \rightarrow \mathbb{C}$, by the absolutely convergent series $\exp(z) := \sum_{n=0}^{\infty}{ \frac{z^n}{n!}}$. It is easy to prove that $\exp$, restricted to the real line, takes real values. A bit more work shows that $\exp$, restricted to the real line, is strictly incresing and everywhere $>0$ and that $\exp(\mathbb{R}) = (0, \infty)$. It follows that there is a bijetive function $\log{}:(0, \infty) \rightarrow \mathbb{R}$, the inverse of $x \mapsto \exp(x)$ from $\mathbb{R} \rightarrow (0, \infty)$. The number $e$ is defined as $e: = \exp(1)$. For any positive real number $a$ and any complex number $z$, one defines $a^z$ by $a^z := \exp(\log(a) z)$. Hence, by definition, $e^z = \exp(z)$ (because $\log{(e)}=1$, by definition) and this makes sense for all complex numbers $z$.
One defines the functions $\cos, \sin : \mathbb{C} \rightarrow \mathbb{R}$ by $\cos(x) := \frac{1}{2}(e^{ix}+ e^{-ix})$ and $\sin(x) := \frac{1}{2i}(e^{ix}- e^{-ix})$. It is a (rather non-trivial) Theorem that there is a unique smallest, strictly positive zero $p$ of the function $\cos$, restricted to the real line. One defines the number $\pi$ by $ \pi := 2p$. It can be shown that the number $\pi$ and the functions $\sin$ and $\cos$ have the "familiar properties". Notice that the identity $e^{ix} = \cos(x) + i \sin(x)$ for $x \in \mathbb{R}$ follows directly from the definitions. As does $e^{i \pi} = -1$. It always strikes me that people find these two identities so amazing.
There can be a bunch of definition of number $e$. Let's stick with this definition: $$ \frac{d}{dt} e^t = e^t. $$
So we want to find what number can be $z(t)=e^{it} =x(t)+iy(t)$: $$ z'(t)=x'(t) + iy'(t)=ie^{it} = ix(t) - y(t),\\ x'(t) = -y(t),\qquad y'(t)=x(t),\\ x''(t) = -y'(t)=-x(t) $$
Finally, $x''(t)+x(t)=0$ and $x(0)=1$, $x'(0)=-y(0)=0$ (since $e^{0i}=e^0=1$), thus $x(t)=\cos t$ as the only function that suffice the equation. Then $y=x'(t)=\sin t$.
So, we have found what number is $e^{it}=\cos t+i\sin t$
Angular Units and Trigonometric Functions
When we talk about arguments to trigonometric functions, there are at least $2$ common angular units: degrees ($360$ to a full rotation) and radians ($2\pi$ to a full rotation). A number of calculators also support gradians ($400$ to a full rotation), which are only used in some countries and usually only in certain occupations (e.g. surveying, mining, geology).
In mathematics, we usually use radians because when angles are measured in radians, we have $$ \lim_{x\to0}\frac{\sin(x)}{x}=\lim_{x\to0}\frac{\tan(x)}{x}=1\tag1 $$ That is, for small angles, $\sin(x)\sim\tan(x)\sim x$. The actual ordering for $|x|\lt\frac\pi2$ is $$ \frac{\sin(x)}{x}\le1\le\frac{\tan(x)}{x}\tag2 $$ Furthermore, when $x$ is in radians, we have the nice series $$ \sin(x)=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}\tag3 $$ and the value of $$ \arctan(x)=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}\tag4 $$ is in radians.
Radians are also natural because an arc which subtends $x$ radians on a circle of radius $r$ has length $rx$.
So when we talk about angles, and don't mention the units, we assume radians.
The Exponential of Imaginary Numbers
For $x\in\mathbb{R}$, we can write $$ e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n\tag5 $$ so it seems reasonable to write $$ e^{ix}=\lim_{n\to\infty}\left(1+\frac{ix}n\right)^n\tag6 $$ Multiplication by $1+\frac{ix}n$ increases the absolute value so minimally, that even when repeated $n$ times, it is insignificant as $n\to\infty$. However, multiplication by $1+\frac{ix}n$ rotates a number on the unit circle by a distance of $\frac xn$ counter-clockwise along the circle. When this is repeated $n$ times, it rotates a number on the unit circle by a distance of $x$.
Thus, $e^{ix}$ is a point on the unit circle whose counter-clockwise distance from $1+0i$ is $x$. This is why we use radians when saying $$ e^{ix}=\cos(x)+i\sin(x)\tag7 $$ To see a more detailed explanation of $(7)$, see this answer.