$(\cosα + i\sinα)^n = \cos(nα) + i\sin(nα)$ for $n$ ,
$(\cosα + i\sinα)^{n+1} = \cos(nα+n) + i\sin(nα+n)$ for $n+1$,
$(\cosα + i\sinα)^{n+1} = \cos(nα)\cos n - \sin(nα).\sinα + i.\sin nα.\cos n +i \cos nα\sin n$
$(\cosα + i\sinα)^{n+1} =\cos nα(\cos(n)+i\sin(n))+\sin(nα)(i\cos n-\sin(n)$
And then I opened $(\cosα + i\sinα)^{n+1}$ like $(\cosα + i\sinα)^n × (\cosα + i\sinα)$ , but I still continue this question but I didn't find exactly. How I can continue this question or new way
Your exposition is a bit unclear. I'll write the inductive step the way I would, then you can take from it what you need. If $(\cos\alpha+i\sin\alpha)^n=\cos n\alpha+i\sin n\alpha$ then$$(\cos\alpha+i\sin\alpha)^{n+1}=(\cos n\alpha+i\sin n\alpha)(\cos\alpha+i\sin\alpha)\\=\cos n\alpha\cos \alpha-\sin n\alpha\sin\alpha+i(\cos n\alpha\sin\alpha+\sin n\alpha\cos\alpha)\\=\cos((n+1)\alpha)+i\sin((n+1)\alpha).$$In this calculation, the first $=$ uses $z^{n+1}=z^nz$, the second uses the definition of multiplication on complex numbers, and the third uses the compound angle formulae for the sine and cosine functions.