algorithm analysis: shows that $ lg (n!) $ is or is not $ o ( lg (n ^ n)) $ and $ omega ( lg (n ^ n)) $

My instructor assigned a problem that asks us to determine which asymptotic limits apply to a given $ f (n) $ for a certain $ g (n) $, in my case $ f (n) = lg (n!) $ Y $ g (n) = lg (n ^ n) $. For clarity, the convention that we use in our class is that $ lg = log_2 $, the "binary logarithm".

I know that by the Stirling approach, $ lg (n!) $ grows in $ O (n lg (n)) $, and evaluating the limit $ lim_ {n to infty} frac {n lg (n)} {n lg (n)} = C $, some constant> 0, and so $ lg (n!) $ is in $ theta ( lg n ^ n) $.

$ theta $ It also means that my $ f (n) $ It's in $ O (g (n)) $ Y $ Omega (g (n)) $, but this does not mean that my $ f (n) $ is in $ o (g (n)) $ or $ omega (g (n)) $.

For that, I think I would have to evaluate $ lim_ {n to infty} frac { lg (n!)} { lg (n ^ n)} $, But I'm not sure.

What strategy would you use to show that? $ f (n) $ is in $ o (g (n)) $ or $ omega (g (n)) $? I would evaluate $ lim_ {n to infty} frac { lg (n!)} { lg (n ^ n)} $?