How to prove that n! have a higher order of growth than n^(log n)? I am aware that n^n have a higher order of growth than n!, but how about n^(log n)? Is there a way to get an alternative form of n^(log n) such that when taking the 
lim n to infinity [alternative form(n^(log n))] / (n!)
it would equate to 0? 

Using a multiplicative variant of Gauss's trick we have:
$$
(n!)^2 = (1 \cdot n) (2 \cdot (n-1)) (3 \cdot (n-2)) \cdots
                ((n-2) \cdot 3) ((n-1) \cdot 2) (n \cdot 1)
                \ge n^n
$$
So
$$
\dfrac{n^{\log n}}{n!}
\le
\dfrac{n^{\log n}}{n^{n/2}}
\le
\dfrac{n^{n/4}}{n^{n/2}}
=
\dfrac{1}{n^{n/4}}
\to 0
$$
because $\log n \le n/4$ for $n$ large enough ($n\ge 9$ actually).
