Using mathematical induction, prove De Moivre's Theorem.

De Moivre's theorem states that (cosø + isinø)ⁿ = cos(nø) + isin(nø). Assuming n = 1 
    (cosø + isinø)¹ = cos(1ø) + isin(1ø)
which is true so correct for n = 1 Assume n = k is true so (cosø + isinø)^k = cos(kø) + isin(kø). Letting n = k + 1 we know that (cosø + isinø)^k+1 = cos((k + 1)ø) + isin((k + 1)ø). But trying to derive the answer from n = k we get:
   (cosø + isinø)^k+1 = (cosø + isinø)^k x (cosø + isinø)      //Using law of indecies
                = (cos(kø) + isin(kø)) x (cosø + isinø) //We've assumed this to be true from n = k
                 = cos(kø)cos(ø) + icos(kø)sin(ø) + isin(kø)cos(ø) - sin(kø)sin(ø) //Expansion
Now using the trigenomitery rules of
sin(a + b) = sin(a)cos(b) + sin(b)cos(a) and cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
and collecting terms to match these formulas we get
                 = cos(kø + ø) + isin(kø + ø)
                = cos((k + 1)ø) + isin((k + 1)ø)  //Taking out ø as a common factor inside the trig functions Which equals our expression gained from De Moivre's theorem for n = k + 1,
and as theorem is true for n = 1 and n = k + 1,
statement is true for all values of n >= 1.