$\mathcal{O}(n^n) > \mathcal{O}(n!) > \mathcal{O}(c^n) > \mathcal{O}(n^c) > \cdots $?

Question

Is the following relationship correct $$\mathcal{O}(n^n) > \mathcal{O}(n!) > \mathcal{O}(c^n) > \mathcal{O}(n^c) > \mathcal{O}(n \cdot Log(n)) > \mathcal{O}( Log(n)) $$

Where $\mathcal{O}$ is big O notation and $c$ is some positive constant.

Answer 1

$n^c$ is only bigger than $n\log n$ when $c>1$.

Answer 2

$O(n\log n) > O(\log n)$. Otherwise, you're fine as far as I can tell.

Answer 3

The last two elements should be swapped, as $n \log n$ grows faster than $\log n$. The other elements are ordered correctly.

Answer 4

Note that it is true (and interesting/useful) that

$\mathcal{O}(n^c) > \mathcal{O}( Log(n))$ for $c>0$, so perhaps the sequence was meant to be:

$\mathcal{O}(n^n) > \mathcal{O}(n!) > \mathcal{O}(b^n) > \mathcal{O}(n^c \cdot Log(n)) > \mathcal{O}(n^c) > \mathcal{O}( Log(n))$, for all $c>0$ and $b>1$.