Proofing de Movire without Induction and in a neat way The "usual way" gone for proving de Movire is via the road of induction. However this road get tiresome and thus wondered, if there were another way. 
However I came up with a proof that relies on some "heavy hitters" and looks far from neat, intuitive, easy to pick up and else.. :
Given the ODE: $y + i y' = 0 $ with $ y(0) = 1 $, one finds $y_1 (x) = e^{ix}   $ and $y_2 (x) = \cos(x) + i \sin(x) $ to be solutions. 
From Picard-Lindelöf it follows: $ y_1 = y_2 $, Euler Identity proven, and then using substitution one gets: 
$ {(e^{ix})^n} = {(\cos(x) + i \sin(x))}^n = {\cos(nx) + i \sin(nx)} =  e^{inx} $, job's done. 
Does someone have a good idea, how to derive de Moivre directly using as little pre-knowledge as possible? Any constructive comment, answer or recommendation is appreciated. 
 A: For me the simplest way is : first you define the exponential as $\exp: \mathbb{C}\to \mathbb{C}$ satisfying $\exp(z) = \sum_{n\geqslant 0} \frac{z^n}{n!}$ (it's very easy to see that this converges for all $z$). The you can show directly by expanding the series that $\exp(z+z') = \exp(z)\exp(z')$, so in particular $\exp(nz) = \exp(z)^n$.
Then you define $\cos$ and $\sin$ by $\cos(z) = \frac{\exp(iz)+\exp(-iz)}{2}$ and $\sin(z) = \frac{\exp(iz)-\exp(-iz)}{2i}$ ; in particular they are defined on $\mathbb{C}$, but if you're only interested in real variables then you can also say $\exp(it) = \cos(t) + i\sin(t)$.
Compared to defining them as solutions of ODE, this has the advantage that you don't need a theorem to show that they exist : you provide formulas (and this is quite useful because the existence of the exponential is crucial in all theorems for existence of solutions to ODE, so this avoids circular contructions).
Then the de Moivre formula is just a rewriting of $\exp(int) = \exp(it)^n$.
A: A very simple way is given in the book by James Uspensky's book titled: Theory of Equations. It is based on multiplication of the polar form representation of a complex number in the argand plane.
Let, there be two complex numbers,  $A = r(\cos\theta + i\sin\theta), B = r_1(\cos\theta_1 + i\sin\theta_1)$.Now, $(\cos\theta + i\sin\theta)(\cos\theta_1 + i\sin\theta_1) =>\cos\theta \cos\theta_1 - \sin\theta \sin\theta_1 +i(\sin\theta \cos\theta_1 + \cos\theta \sin\theta_1) \implies AB = rr_1[\cos(\theta + \theta_1) + \sin(\theta + \theta_1)]$
=> For the case of taking $n$ angles, all equal, & the modulus of all being $1$, i.e. $\theta_1= \theta_2=...=\theta_n=\theta$, and $r_1=r_2=...=r_n=1$ this formula gives the needed result, i.e. the de Moivre's formula:
$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$
But, I feel the generalization part (taking $n$ complex quantities) implicitly uses induction (weak case) only,
