TITLE: Last 5 digits of factorial sum
QUESTION [5 upvotes]: Find the last 5 digits of
  $$1!+2\cdot2!+3\cdot3!+\cdots+23\cdot23!+24\cdot24!$$

I have no idea how to find an elegant solution to this since the "last 5 digits" part makes it much harder. Usually the number of fives in each factorial reduces the computation here, but here it doesn't work.

REPLY [13 votes]: Hints: 
$n\times n!=(n+1)!-n!$, so the sum telescopes to $25!-1$.
$25!$ has $6$ factors of $5$ and more than that many of $2$.