Notice that k! mod 10 = 0 for all k=5,6,..,100 since all these factorials include at least 2 and 5 as factors. So, you need only to compute 1-2+3!-4!=-19. Moreover, 1!-2!<0, 3!-4!<0, so the whole sum is negative too, hence 9 is a unit digit of 1!-2!+3!-4!+... -98!+99!-100!