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 $\mathcal{o}(g(n)) = \{f(n)|\forall C &gt; 0 \exists n_{0} \forall n\geq n_{0}:f(n) &lt; C * g(n)\}$. 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 $C &gt; 0$, 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 $2^{n} &lt; n!, \forall n\geq 4$. 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.)
Pick $n_0 = \max\{ -\log c, 10 \}$. Then, for all $n \ge n_0 \ge 10$: 
$$
2^n = \frac{4^n}{2^n} \le \frac{4^n}{2^{n_0}} \le \frac{4^n}{1/c} =
c \cdot (4^{n-10} \cdot 4^{10}) &lt;
c \cdot (4^{n-10} \cdot 10!) \le cn!
$$