TITLE: Bijective proof of weak form of Stirling's approximation
QUESTION [6 upvotes]: There are short and sweet proofs of various forms of Stirling's approximation. But even the sweetest among them don't instill the same conviction in the reader as a direct bijective proof.
Computer scientists can often get away with a very weak form of Stirling's approximation:
$$(n/2)^{n/2} \leq n! \leq n^n$$
From this follows
$$(n/2)\ log(n) - (n/2)\ log(2) \leq log(n!) \leq n\ log(n)$$
and therefore $log(n!) = \Theta(n\ log(n))$. This is sufficient to establish the lower bound on comparison-based sorting algorithms and many other asymptotic bounds.

Does anyone know a natural bijective proof of $(n/2)^{n/2} \leq n! \leq n^n$?

REPLY [6 votes]: Think of all maps from the first $n/2$ elements of {$1,...,n$} to the last $n/2$.  Say, let $a_1< \ldots < a_k \to z$.  Make a cycle $a_1 \to a_2 \to \ldots \to a_k \to z \to a_1$.  Do this for all $z$.  The details are straightforward.  This proves the lower bound.