I am looking at the following exercise:
Show with induction, that the $i^{\rm th}$ Fibonacci number satisfies the equality:
$$F_i=\frac{\phi^i-\hat{\phi}^i}{\sqrt{5}}$$ where $\phi$ represents the golden ratio and $\hat{\phi}$ is its conjugate.
It is known that: $$F_0=0 \\ F_1=1 \\ F_i=F_{i-1}+F_{i-2}, i \geq 2 \\ \phi=\frac{1+ \sqrt{5}}{2}, \hat{\phi}=\frac{1-\sqrt{5}}{2}$$
That's what I have tried:
$$F_0=\frac{\phi^0-\hat{\phi}^0}{ \sqrt{5}}=0 \checkmark$$
$$F_1=\frac{\phi-\hat{\phi}}{\sqrt{5}}=\frac{\frac{1+\sqrt{5}}{2}-\frac{1-\sqrt{5}}{2}}{\sqrt{5}}=1 \checkmark$$
- We suppose that the relation stands for $2 \leq i \leq n$: $$F_i=\frac{\phi^i-\hat{\phi}^i}{\sqrt{5}}$$
- We will show that the relation stands for $n+1$:
$$F_{n+1}=F_n+F_{n-1}=\frac{\phi^n-\hat{\phi}^n}{ \sqrt{5}}+\frac{\phi^{n-1}-\hat{\phi}^{n-1}}{ \sqrt{5}}=\frac{\phi^{n-1}(\phi+1)-\hat{\phi}^{n-1}(\hat{\phi}+1)}{\sqrt{5}} (*)$$
Then,using the equality $\phi^2=\phi+1$,I got:
$$(*)=\frac{\phi^{n-1} \phi^2-\hat{\phi}^{n-1} \hat{\phi}^2}{\sqrt{5}}=\frac{\phi^{n+1}-\hat{\phi}^{n+1}}{\sqrt{5}}$$
Can I just use the equality $\phi^2=\phi+1$ without a proof, or do I have to prove it?