Let $$\begin{array}{} (\cos x + i \sin y)^n &= \cos nx + i \sin nx & (1)\\ (\cos x + i \sin y)^{n+1} &= \cos (n+1)x + i \sin (n+1)y &(2) \end{array}$$ Then (1) $\times (\cos x + i\sin y)$ - (2) $\implies \sin x \sin nx = \sin y \sin ny$. If $x\ne y$, then without loss of generality, let's assume $x < y$. it follows that $\sin x < \sin y < 1 \implies |\sin nx | \ge |\sin ny|$ Taking absolute values on (1) $\implies (\cos^2 x + \sin^2 y)^n = \cos^2 nx + \sin^2 ny$. Hence $1=(\cos^2 x +\sin^2 x)^n < (\cos^2x + \sin^2 y)^2 = \cos^2 nx + \sin^2ny \le \cos^2nx + \sin^2 nx=1$. This is a contradiction. Therefore $x=y$ must hold. In this case $(\cos x+i\sin y)^n=\cos nx + i\sin ny$ will always hold.