Fibonacci sequence: how to prove that $\alpha^n=\alpha\cdot F_n + F_{n-1}$?

It's easier to prove this by induction. Note that $F_1\alpha + F_0=\alpha+0=\alpha^1$.

Then use that $\alpha$ is a root of $x^2-x-1=0$ to show that if $\alpha^n=F_n\alpha+F_{n-1}$ then it follows that $\alpha^{n+1}=F_{n+1}\alpha + F_{n}$.

First, for $n=1$, $$ \begin{align} \alpha^1 &=\alpha F_1+F_0\\ &=\alpha\cdot1+0\\ &=\alpha \end{align} $$ Suppose it's true for $n$, then because $\alpha^2=\alpha+1$, $$ \begin{align} \alpha^{n+1} &=\alpha\cdot\alpha^n\\ &=\alpha(\alpha F_n+F_{n-1})\\ &=\alpha^2F_n+\alpha F_{n-1}\\ &=(\alpha+1)F_n+\alpha F_{n-1}\\ &=\alpha(F_n+F_{n-1})+F_n\\ &=\alpha F_{n+1}+F_n \end{align} $$ Thus, it is true for $n+1$. By induction, the identity is true for all $n\ge1$.

Using Euler-Binet Formula,

$$F_{n+1}=\frac{\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta}$$

$$=\frac{\alpha^n(\alpha-\beta)+\beta\cdot \alpha^n-\beta^{n+1}}{\alpha-\beta}$$

$$=\alpha^n+\beta\cdot \frac{\alpha^n-\beta^n}{\alpha-\beta}$$

$$=\alpha^n+\beta\cdot F_n =\alpha^n-\frac{F_n}\alpha\text{ as }\alpha\cdot \beta=-1$$

$$\implies \alpha\cdot F_{n+1}+F_n=\alpha^{n+1} $$