Prove De Moivre's by induction for the positive integers

First we check the formula for our first case, n=1 where (cosx + isinx)ⁿ =cosnx + isinnx (cosx + isinx)¹ = cos(1x) + isin(1x) - holds true for n=1 next we assume true for our case n=k(cosx + isinx)^k = coskx + isinkxnext we show that if the case for n=k is true, then the case n=k+1 is true.by inputting n=k+1 we get (cosx+isinx)^(k+1)=(cosx+isinx)(cosx+isinx)^know as we assumed for n=k ,this equals (cosx+isinx)(coskx+isinkx)this gives cosxcoskx +icosxsinkx +isinxcoskx -sinxsinkxsimplifying to (coskxcosx -sinxsinkx) + i(cosxsinkx + sinxcoskx)finally through the compound angle formulae we reach our desired result: (cosx+isinx)^k+1 = cos((k+1)x) + isin((k+1)x) conclusion: If n=k holds true, then n=k+1 holds true. Since n=1 holds true we have now shown that De Moivre's theorem holds true for all positive integers.