From what I've been taught, to compare the growth orders of two functions, $t(n)$ and $s(n)$, we find $$ lim_{n\to\infty} \frac{t(n)}{s(n)} $$ where $n$ is the input size.
- If the limit is zero, then $s(n)$'s order of growth is greater than $t(n)$'s.
- If the limit is a constant, then $s(n)$ and $t(n)$ are in the same growth order/efficiency class.
- If the limit approaches infinity, then $t(n)$'s growth order is greater than greater than $s(n)$'s.
So for $(n-2)!$ vs $n!$ we have: $$lim_{n\to\infty} \frac{(n-2)!}{n!} =lim_{n\to\infty} \frac{(n-2)!}{(n)(n-1)(n-2)!} = lim_{n\to\infty} \frac{1}{(n)(n-1)} = 0 $$
This seems very counter-intuitive, because at first glance I'd think that $(n-2)!$ would have the same growth order as $n!$. Did I go wrong somewhere, or is it actually in a growth order of its own?