TITLE: Proof that $\sum\limits_{i=1}^k \log(i)$ belongs to $O(k)$
QUESTION [5 upvotes]: I'm studying time complexity of binomial heaps and there's one operation (the make-heap operation) that does not make sense to me unless the following is true.
$\sum\limits_{i=1}^k \log(i)$ belongs to $O(k)$
Please help me find a proof for that statement.
Any help appreciated.

REPLY [9 votes]: Elvis's answer is nicer than this, but since the question comes from intro algorithms, I'd point out that for many elementary CS applications the trivial bounds like:
$$
(n/2)\log(n/2) = (n/2)(\log n - 1) \le \sum_{i=1}^n \log i\le n\log n
$$
are good enough and worth trying if you're just doing homework.
Update: The lower bound here can be obtained by noticing that half of the terms in the sum are at least $\log(n/2)$, since log is monotone.  Also, fixed a dropped set of brackets.