Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 1 / 9F / c / Solution

Applying the functional equation at $x^{{\alpha }^{-1}}$ repeatedly givesSince $x^{{\alpha }^{-n}}\to 1$, continuity at $1$ gives $f(x)=f(1)$. Thus $f$ is constant.

Without continuity the claim is false. The function $f(1)=1$ and $f(x)=0$ for $x\ne 1$ satisfies $f(x)=f(x^{\alpha })$ because $x^{\alpha }=1$ for positive $x$ exactly when $x=1$, but it is not constant.