Proof by induction on n.

Base:\(n = 0\): \((\cos(x) + i \sin(x))^0 = 1\) and \(\cos(0x) + i \sin(0x) = 1 + i0 = 1\). The two are equal and so the claim is true for the base case.

Inductive hypothesis:Suppose that \((\cos(x) + i \sin(x))^n = \cos(nx) + i \sin(nx)\) holds for \(n = 0,1, \ldots, k\).

Rest of the inductive step:By the inductive hypothesis, we know that \((\cos(x) + i \sin(x))^k = \cos(kx) + i \sin(kx)\).Also recall from high school trignometry that $$\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)$$ $$\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)$$

Using these equations, we can compute

\( \begin{eqnarray*} (\cos(x) + i \sin(x))^{k+1} &=& (\cos(x) + i \sin(x))^{k}(\cos(x) + i\sin(x)) \\ &=& (\cos(x) + i \sin(x))\cdot(\cos(kx) + i \sin(kx)) \\ &=& \cos(x)\cos(kx) + i\cos(x)\sin(kx) + i\sin(x)\cos(kx) - \sin(x)\sin(kx) \\ &=& (\cos(x)\cos(kx) - \sin(x)\sin(kx)) + i(\sin(kx)\cos(x) + \sin(x)\cos(kx)) \\ &=& \cos(x + kx) + i\sin(x + kx) \\ &=& \cos((k+1)x) + i\sin((k+1)x) \end{eqnarray*} \)

So \((\cos(x) + i \sin(x))^{k+1} = \cos((k+1)x) + i\sin((k+1)x)\) which is what we needed to show.

Notice the use of the word "recall" in the inductive step. The word "recall" is often used to politely work around the fact that some of the audience has forgotten the facts in question. It's important to remind the readers of the equations, because one of the later steps will seem completely mysterious if they don't recall the equations. But it would be impolite to suggest too strongly that they've forgotten a fact that everyone learned in high school.