De Moivre's formula by induction

Base case. $n=1$: trivial.

Inductive step. Assume the formula for $n$. Then
$$(\cos\theta+i\sin\theta)^{n+1}=(\cos n\theta+i\sin n\theta)(\cos\theta+i\sin\theta).$$
Expanding:
$$=\cos n\theta\cos\theta-\sin n\theta\sin\theta+i(\sin n\theta\cos\theta+\cos n\theta\sin\theta).$$
Using the addition formulas:
$$=\cos((n+1)\theta)+i\sin((n+1)\theta).$$

Conclusion. The formula holds for all $n\ge1$. (It extends to all $n\in\mathbb{Z}$ using $(\cos\theta+i\sin\theta)^{-1}=\cos(-\theta)+i\sin(-\theta)$.)