Let $x,y \in \left(0, \dfrac{\pi}{2}\right)$. Suppose the equality $$(\cos{x}+i\sin{y})^n = \cos{nx}+i\sin{ny}$$ holds for two consecutive positive integers $n$ and $n+1$. Prove it holds for all positive integers $n$.
I tried to write the relation for both $n$ and $n+1$, but the computations were going like this: $$\begin{align} (\cos{x}+i\sin{y})^n (\cos{x}+i\sin{y}) &= (\cos{nx}+i\sin{ny})(\cos{x}+i\sin{y}) \\ &= (\cos{(n+1)x}+i\sin{(n+1)y}) \end{align}$$ Then I tried to apply the equality between two complex numbers. But the computations were very ugly.
Please help!
We'll begin by proving the phase is as desired. Let$$\theta:=\operatorname{atan2}(\sin y,\,\cos x),\,t_k:=\tan k\theta,\,c_k:=\cos kx,\,s_k:=\sin ky$$so $t_n=\frac{s_n}{c_n},\,t_{n+1}=\frac{s_{n+1}}{c_{n+1}}$. (We can actually replace $\operatorname{atan2}$ with $\arctan$, since $x,\,y$ are acute.) The identity$$\tan(2A-B)=\frac{\tan 2A-\tan B}{1+\tan 2A\tan B}=\frac{2\tan A-\tan B+\tan^2A\tan B}{1-\tan^2A+2\tan A\tan B}$$with $A:=(n+1)\theta,\,B:=n\theta$ implies$$t_{n+2}=\frac{2t_{n+1}-t_n+t_{n+1}^2t_n}{1-t_{n+1}^2+2t_{n+1}t_n}=\frac{2s_{n+1}c_{n+1}c_n-s_nc_{n+1}^2+s_{n+1}^2s_n}{c_{n+1}^2c_n-s_{n+1}^2c_n+2s_{n+1}s_nc_{n+1}}=\frac{s_{n+2}}{c_{n+2}},$$where the last $=$ follows from a similar use of the identities$$\begin{align}\sin(2A-B)&=2\sin A\cos A\cos B-\sin^2A\sin B+\cos^2A\sin B,\\\cos(2A-B)&=\cos^2A\cos B-\sin^2A\cos B+2\sin A\cos A\sin B,\end{align}$$and the inductive step in the proof that $k\ge n\implies t_k=\frac{s_k}{c_k}$ is complete. We can go backwards in a similar way, obtaining the $k=n-1$ case by transposing $A,\,B$ in the above logic, and so we can induct in both directions.
Now we just need to check the modulus. Since the results above about phases remain true with $x,\,y,\,\theta\mapsto kx,\,ky,\,k\theta$ for $k\in\Bbb N$, $$k\operatorname{atan2}(\sin y,\,\cos x)=\operatorname{atan2}(\sin ky,\,\cos kx)$$for all such $k$. Hence $x=y$, so $|c_k+is_y|=1$.