Fibonacci Number greater than Golden Section to Power less Two

Theorem
For all $n \in \N_{\ge 2}$:
 * $F_n \ge \phi^{n - 2}$

where:
 * $F_n$ is the $n$th Fibonacci number
 * $\phi$ is the golden section: $\phi = \dfrac {1 + \sqrt 5} 2$

Proof
The proof proceeds by induction.

For all $n \in \N_{\ge 2}$, let $P \left({n}\right)$ be the proposition:
 * $F_n \ge \phi^{n - 2}$

Basis for the Induction
$P \left({2}\right)$ is true, as this just says:
 * $F_2 = 1 = \phi^0 = \phi^{2 - 2}$

It is also necessary to demonstrate $P \left({3}\right)$ is true:
 * $F_3 = 2 \ge \dfrac {1 + \sqrt 5} 2 = \phi = \phi^{3 - 1}$

This is the basis for the induction.

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

So this is the induction hypothesis:
 * $F_k \ge \phi^{k - 2}$

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

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:
 * $\forall n \in \N_{\ge 2}: F_n \ge \phi^{n - 2}$