Showing that $\lg(n!)$ is or is not $o(\lg(n^n))$ and $\omega(\lg(n^n))$
You seem tot be trying to prove something that is false. If $f=O(g)$ then $\lim_{n\to\infty}g/f &gt; 0$ so $f\neq \omega(g)$. Similarly, if $f=\Omega(g)$ then $f\neq o(g)$.

Since you already have that $\lg n! = \Theta(\ln n^n)$, that gives you big-$O$ and big-$\Omega$, which preclude little-$\omega$ and little-$o$, respectively.