I need to prove
$(n+1)!=\omega (2^n)$
and
$(n-1)!=o(n^n)$
From what I understand, the small omega and o signify, that these are loose upper and lower bounds. If I'm not wrong, this should mean $(n+1)!\geq c(2^n)\forall c>0, n>n_0, \exists n_0>0$ and $(n-1)!\leq cn^n\forall c>0, n>n_0, \exists n_0>0$
However, I do not know how to approach a proof for all values of c. What happens if c approaches infinity or 0? How do I go about finding a value of $n_0$?
As is written in comment, limit variant of definition for little-$o$ gives $$\lim\limits_{n \to \infty}\frac{(n-1)!}{n^n}=0$$ same way with little-$\omega$ gives $$\lim\limits_{n \to \infty}\frac{(n+1)!}{2^n}=\infty$$ From here you can found also values for $n_0$. For example, can be used inequalities:
$$\left(\frac{n}{e}\right)^n \lt n! \lt e \left(\frac{n}{2}\right)^n$$