Recently I ran across a derivation of the exponential generating function for the derangement numbers that I like better than the ones I had seen before. (Perhaps it’s standard, but it was new to me.) It’s Problem 10 in Chapter 5 of the text  I’m currently using for combinatorics.
A derangement is a permutation with no fixed points. One of the standard recurrences for the number of derangements on n elements is , valid for . The base case is . Now, multiply this recurrence by and sum over n to get
(I’ve also given this answer on math.SE.)
- W. D. Wallis and J. C. George, Introduction to Combinatorics, CRC Press, 2011.