Is n!=O(2^(n lg n))?
Remember, big-O is essentially an upper bound so we do have
$$
n! = O(n^n)
$$
in the sense that there is a constant $c$ for which $n!\le c\;n^n$ eventually. [Note, by the way, that there's no function $T(\cdot)$ involved here.] 

Now let's see if $n!=O(2^{n\log n})$. We can argue this way (assuming the log is base-2):
$$
2^{n\log n} = 2^{(\log n) n} = \left(2^{\log n}\right)^n = n^n
$$
so we can conclude that the two sets $O(n^n)$ and $O(2^{n\log n})$ are in fact equal.