I have been searching for a while now but couldn’t find anything about this exact pair of functions with the little $ \mathcal{o}$ notation.

Given the functions $ f(n) = 2^{n}$ and $ g(n) = n!$ I am supposed to prove, or disprove, the following statement: $ f(n) \in \mathcal{o}(g(n))$ .

I am fairly sure that it’s true but now I need an idea of how to show this. We have just started out with this whole concept and this is the second exercise, the first one being a relatively easy big $ \mathcal{O}$ task. But this exercise is just beyond me right now. The only definition I am allowed to use (meaning: NO LIMITS) is $ \mathcal{o}(g(n)) = \{f(n)|\forall C > 0 \exists n_{0} \forall n\geq n_{0}:f(n) < C * g(n)\}$ . This means other than with big $ \mathcal{O}$ , where it suffices to show that there’s at least one pair $ C$ and a $ n_{0}$ so that $ f(n) \leq C * g(n)$ $ \forall n \geq n_{0}$ , I now have to prove that for **every** $ C > 0$ , there is such a $ n_{0}$ so that the condition stated in the set is true.

I first have been thinking about the functions, and I would have an answer for $ \mathcal{O}$ , because you can prove with induction that $ 2^{n} < n!, \forall n\geq 4$ . Meaning my C would be 1 here. However, I have no idea how to prove it for every C and would be grateful for any guidance! (It would already help to know how to start. Probably like, let $ C$ be greater 0, and then I have to show that for any Value of this $ C$ , there is… because… My biggest struggle is to find meaningful estimations to get a chain of inequalities.)