Proving that $\alpha^{n-2}\leq F_n\leq \alpha^{n-1}$ for all $n\geq 1$, where $\alpha$ is the golden ratio I got stuck on this exercise. It is Theorem 1.15 on page 14 of Robbins' Beginning Number Theory, 2nd edition.

Theorem 1.15. $\alpha^{n-2}\leq F_n\leq \alpha^{n-1}$ for all $n\geq 1$.
Proof: Exercise.

(image of relevant page)

Using Binet's Fibonacci Number Formula,
$\alpha+\beta=1,\alpha\beta=-1$ and $\beta<0$
\begin{align}
F_n-\alpha^{n-1}=
&
\frac{\alpha^n-\beta^n}{\alpha-\beta}-\alpha^{n-1}
\\
=
&
\frac{\alpha^n-\beta^n-(\alpha-\beta)\alpha^{n-1}}{\alpha-\beta}
\\
=
&
\beta\frac{(\alpha^{n-1}-\beta^{n-1})}{\alpha-\beta}
\\
=
&
\beta\cdot F_{n-1}\le 0 
\end{align}
 if $n\ge 1$.
\begin{align}
F_n-\alpha^{n-1}=
&
\frac{\alpha^n-\beta^n}{\alpha-\beta}-\alpha^{n-2}
\\
=
&
\frac{\alpha^{n-1}\cdot \alpha-\beta^n-\alpha^{n-1}+\beta\cdot \alpha^{n-2}}{\alpha-\beta}
\\
=
&
\frac{\alpha^{n-1}\cdot (1-\beta)-\beta^n-\alpha^{n-1}+\beta\cdot \alpha^{n-2}}{\alpha-\beta}
\\
=
&
\frac{\beta(\alpha^{n-2}-\alpha^{n-1})-\beta^n}{\alpha-\beta}
\\
=
&\frac{\beta\cdot \alpha^{n-2}(1-\alpha)-\beta^n}{\alpha-\beta}
\\
=
&
\frac{\beta\cdot \alpha^{n-2}(\beta)-\beta^n}{\alpha-\beta}
\\
=
&
\beta^2\frac{\alpha^{n-2}-\beta^{n-2}}{\alpha-\beta}
\\
=
&
\beta^2F_{n-2}\ge 0
\end{align}
if $n\ge 1$.
