Sequence of Powers of Golden Mean

Theorem
Consider the golden mean $\phi = 1 \cdotp 618 \ldots$

and in general:
 * $\phi^n = F_n \phi + F_{n - 1}$

where $F_n$ is the $n$th Fibonacci number.

Proof
The proof proceeds by strong induction.

For all $n \in \Z_{\ge 1}$, let $P \left({n}\right)$ be the proposition:
 * $\phi^n = F_n \phi + F_{n - 1}$

$P \left({1}\right)$ is the case:

Thus $P \left({1}\right)$ is seen to hold.

Basis for the Induction
$P \left({2}\right)$ is the case:

Thus $P \left({2}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({j}\right)$ is true, for all $j$ such that $0 \le j \le k$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is the induction hypothesis:
 * $\phi^k = F_k \phi + F_{k - 1}$

from which it is to be shown that:
 * $\phi^{k + 1} = F_{k + 1} \phi + F_k$

Induction Step
This is the induction step:

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Second Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \Z_{>0}: \phi^n = F_n \phi + F_{n - 1}$