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?

$$
F_{k+1} = F_k + F_{k-1} \le k! + (k - 1)! \le k! + k! \le 2 k! \le (k + 1) k!
$$
