The definition of Big-O is $$u_n = O(v_n) \iff (\exists c \in \mathbb{R}^{*})\,\, (\exists N \in \mathbb{N}) \,\, n > N \implies u_n < c \, v_n$$
Based on that I am trying to find the upper bound for this function $log(n!)/log(n)^3$.
What I've done so far is:
$log(n!) = log(1) + log(2) + ... + log(n!) <= n*log(n) \implies log(n!) = O(n*logn) $
So, $log(n!)/log(n)^3 = O(n*log(n)/log(n)^3)$.
Can we simplify this even more by saying that: $O(n*log(n)/log(n)^3) = O(n)$ ?