There actually an elementary way to derive everything from series definitions of the exponential and trigonometric functions:
First obtain the series for $\exp$ on $\mathbb{C}$ about $0$ where $\exp$ is defined as a function such that $\exp' = \exp$ and $\exp(0) = 1$, and prove that the series converges on all of $\mathbb{C}$. Then for any $w \in \mathbb{C}$ find the series for $z \mapsto \exp(z+w)$ about $0$, which will be just $\exp(w)$ times the series for $\exp$, and thus it also converges on all of $\mathbb{C}$ and so $\exp(z+w) = \exp(z) \exp(w)$ for any $z,w \in \mathbb{C}$.
Next prove that the series for $\cos$ converges on $\mathbb{R}$, and likewise the series for $\sin$. Then note that $\exp(ix) = \cos(x) + i \sin(x)$ for any real $x$ because of the series. $\exp(i) \exp(-i) = \exp(0) = 1$ and $\exp(-i) = \cos(-1) + i \sin(-1) = \cos(1) - i \sin(1) = \exp(i)^*$ ("$^*$" denotes conjugation), and so $|\exp(i)| = 1$. Now we easily get $|\exp(ik)| = |\exp(i)^k| = 1^k = 1$ for any integer $k$, and $|\exp(i\frac{k}{m})|^m = |\exp(i\frac{k}{m})^m| = |\exp(ik)| = 1$ implies $|\exp(i\frac{k}{m})| = 1$ for any integers $k,m$. Since $\exp$ is continuous, $|\exp(ix)| = 1$ for any real $x$.
Finally we define $π$ as twice the first positive root of $\cos$, which we know must exist in $(0,2)$ by intermediate value theorem because we have $\cos(x) < 1 - \frac{x^2}{2!} + \frac{x^4}{4!}$ for any $x$ by repeated differentiation and $\cos(0) = 1 > 0$ and $\cos(2) < 1 - \frac{2^2}{2!} + \frac{2^4}{4!} = -\frac{1}{3} < 0$. Similarly we have $\sin(x) > x - \frac{x^3}{3!} > 0$ for any $x \in (0,2)$. These imply $\exp(i\frac{π}{2}) = \cos(\frac{π}{2}) + i \sin(\frac{π}{2}) = 0 + i(1)$, because the unit circle intersects the imaginary axis only at $i,-i$, and $\sin(\frac{π}{2}) > 0$. Hence we get $\exp(i2π) = \exp(i\frac{π}{2})^4 = i^4 = 1$. Marvelous! We're done!
Personally I like this way because it reveals the geometric nature of $\exp$ nicely and completely naturally, but I don't know why none of my teachers ever taught me this way.
