# Prove little o with just the definition 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.) 