The inductive step is $$\frac{1}{\sqrt{5}}\phi^{n-1}+\frac{1}{\sqrt{5}}\phi^{n-2}=\frac{1}{\sqrt{5}}\phi^{n-2}(\phi-1)=\frac{1}{\sqrt{5}}\phi^n$$, because $$\phi^2=\phi+1$$, and all that is true for $$\psi$$ too. The formula in the book is correct. $$\psi<0$$, so its integer powers are very small for large $$n$$.