How to prove De Moivre's theorem inductively It is given that $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ where $n\in Z^+$.
I can show it works for $n=1$ but I am stuck in showing it inductively. I have got as far as below but are stuck in the rearranging:
$$(\cos\theta + i\sin\theta)^{k+1}= (\cos k\theta + i\sin k \theta)(\cos\theta + i\sin\theta)$$
How can I rearrange the left to look like the right?
 A: $$(\cos k\theta+i\sin k\theta)(\cos\theta+i\sin\theta)=\\
\cos k\theta\cos\theta-\sin k\theta\sin\theta+i(\cos k\theta\sin\theta+\sin k\theta\cos\theta)\\
=\cos(k+1)\theta+i\sin(k+1)\theta
 $$
A: $$(\cos \theta + i\sin \theta)^k = \cos k\theta + i\sin k\theta$$
$$\text{So for } (\cos \theta + i\sin \theta)^{k+1} = (\cos \theta + i\sin \theta)^k \cdot (\cos \theta + i\sin \theta) = (\cos k\theta + i\sin k\theta)\cdot (\cos \theta + i\sin \theta)$$
$$=\cos k\theta\cos\theta +i\sin\theta\cos k\theta +i\sin k\theta\cos\theta +i^2\sin{k\theta}\sin\theta$$
$$=(\cos{k\theta}\cos\theta - \sin k\theta\sin\theta)+i(\sin\theta\cos k\theta +\sin k\theta\cos\theta)$$
$$=\cos(k+1)\theta + i\sin(k+1)\theta$$
Hence, this proves that $(\cos\theta +i\sin\theta)^n = \cos n\theta + i\sin n\theta$.
A: By induction,
$(\cos \theta + i\sin \theta )^{k+1} = (\cos \theta + i\sin \theta )^k.(\cos \theta + i\sin \theta ) = (\cos k\theta + i\sin k\theta )(\cos \theta + i\sin \theta )$
$(\cos \theta + i\sin \theta )^{k+1} = \cos k\theta\cos \theta + i\sin k\theta\cos \theta + i\sin \theta \cos k\theta -  \sin k\theta\sin \theta$
$(\cos \theta + i\sin \theta )^{k+1} = \cos((k+1)\theta) + i\sin((k+1)\theta) $
For $k=n \in \mathbb{N}$
$$(\cos \theta + i\sin \theta )^n = \cos n\theta + i\sin n\theta$$
We can also prove this by simple exponentiation rule and Euler's form of complex numbers.
$z = \cos \theta + i\sin \theta = e^{i\theta}$
As, $e^{ix} = \cos x+ i\sin x\cdots(i)$
$z^n = (e^{i\theta})^n$ 
As, $(e^a)^b= e^{ab}$
$z^n = (e^{i\theta})^n =e^{in\theta}$
From (i), we can write,
$$z^n = (\cos \theta + i\sin \theta )^n = \cos n\theta + i\sin n\theta$$

A: $$(\cos\theta + i\sin\theta)^{k+1}=(\cos\theta + i\sin\theta)^{k}(\cos\theta + i\sin\theta)=(\cos k\theta + i\sin k \theta)(\cos\theta + i\sin\theta)$$
