I'm trying to get a tight upper bound for the following expression:
${{n \log n} \choose {\log n}}\frac{1}{n ^ {\log n}}$
The bound that I've obtained is too loose, which is as follows:
$ m = nM, \\ M = \log n$
${{m} \choose {M}}\frac{1}{n ^ {M}} = \frac{({m} )!}{(m - M)! M!} \frac{1}{n^M} = \frac{m \cdot (m -1) \cdot (m - 2) \cdots (m - (M - 1))}{n \cdot n \cdot n \cdots n}\frac{1}{M!} \leq (\frac{m}{n})^M \cdot \frac{1}{M!} = \frac{M^M}{M!}$
But,the bound on the RHS is far greater than the LHS, as I can see when I graph the two quantities (Red is LHS, Blue is RHS):
Any help in how to get a tighter bound would be much appreciated!
EDIT: The corrected graph of RHS vs LHS is added, which turns out to be a tight bound after all