Last digit of factorial

999! has more than 200 zeroes. Because 999!=1×2×3×4×5................

2×5=10. Hence the last digit of 999! is 0

• 1!=1
• 2!=2
• 3!=6
• 4!=24
• 5!=120
• from then on the unit digit is 0 therefore 0.

Note that $5!$ has a units digit of $0$ , so $n!$ has a units digit of $0$ , for all integral values of $n$ at least $5$ , so our answer is $\boxed {0}$ .

0 times anything is 0 and 5 times 2 equals 10 which ends with zero, so any number times 10 is going to end up with units digit of 0.