# How to prove that $\mathrm{Fibonacci}(n) \leq n!$, for $n\geq 0$

I am trying to prove it by induction, but I’m stuck
$$\mathrm{fib}(0) = 0 < 0! = 1;$$
$$\mathrm{fib}(1) = 1 = 1! = 1;$$

Base case n = 2,

$$\mathrm{fib}(2) = 1 < 2! = 2;$$

Inductive case assume that it is true for (k+1) $k$
Try to prove that $\mathrm{fib}(k+1) \leq(k+1)!$

$$\mathrm{fib}(k+1) = \mathrm{fib}(k) + \mathrm{fib}(k-1) \qquad(LHS)$$

$$(k+1)! = (k+1) \times k \times (k-1) \times \cdots \times 1 = (k+1) \times k! \qquad(RHS)$$

……

How to prove it?