It would take a lot of pencils to work this out by hand, but an insight makes it easy: For any n ≥ 5, n! must have the factors 2 and 5 and hence be a multiple of 10, with zero as a final digit. So to solve the puzzle we only have to work out 1! + 2! + 3! + 4! = 33. The last digit is 3.
(“A Property of Large Factorial Numbers,” Mathematical Gazette 81:490 [March 1997], 80-81.)