Is $\sin(n^k) ≠ (\sin n)^k$ in general?
Solution 1:
If $\sin(n^k)=\sin(n)^k$, then $\frac{e^{in^k}-e^{-in^k}}{2i}=(\frac{e^{in}-e^{-in}}{2i})^k$, so that $e^i$ would be algebraic.
But that directly contradicts the Lindemann-Weierstrass theorem because $i$ is algebraic.