Algorithmic Analysis – Show that $ lg (n!) $ Is or is not $ o ( lg (n ^ n)) $ and $ omega ( lg (n ^ n)) $

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

I know that by approximation of Stirling, $ lg (n!) $ grows up in $ O (n lg (n)) $and evaluate the limit $ lim_ {n to infty} frac {n lg (n)} {n lg (n)} = C $, a constant> 0, etc. $ lg (n!) $ is in $ theta ( lg n ^ n) $.

$ theta $ also means that my $ f (n) $ is $ O (g (n)) $ and $ Omega (g (n)) $but that does not mean that my $ f (n) $ is in $ o (g (n)) $ or $ omega (g (n)) $.

For that, I think I should evaluate $ lim_ {n to infty} slash { lg (n!)} { lg (n ^ n)} $but I'm not sure.

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