Prove little o with just the definition
Pick $n_0 = \max\{ -\log c, 10 \}$. Then, for all $n \ge n_0 \ge 10$: 
$$
2^n = \frac{4^n}{2^n} \le \frac{4^n}{2^{n_0}} \le \frac{4^n}{1/c} =
c \cdot (4^{n-10} \cdot 4^{10}) &lt;
c \cdot (4^{n-10} \cdot 10!) \le cn!
$$