Problem: n = 1 ! + 2 ! + 3 ! . . . . . + 10 ! . what is the last digit of n ^ n ?
Rationale: as you see 5 ! till 10 ! each unit digit is zero . so 1 ! + 2 ! + 3 ! + 4 ! = 33 so unit digit 3 + 0 = 3 n = 3 n ^ n = 3 ^ 3 = 27 so last digit is 7 . answer : e
annotated_formula: power(subtract(10, 3), const_1)