Factorial

0! is a special case that is explicitly defined to be 1

1! = 1

2! = 2 * 1 = 2

3! = 3 * 2 * 1 = 6

4! = 4 * 3 * 2 * 1 = 24

5! = 5 * 4 * 3 * 2 * 1 = 120

6! = 6 * 5 * 4 * 3 * 2 * 1 = 720

7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040

8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 40320

9! = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 362880

10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 3628800

etc.

Notice that

4! + 1 = 24 + 1 = 25 = 5^2

5! + 1 = 120 + 1 = 121 = 11^2

But

6! + 1 = 720 + 1 = 721 is not a square number

7! + 1 = 5040 + 1 = 5041 = 71^2

8! + 1 = 40320 + 1 = 40321 is not a square number

9! + 1 = 362880 + 1 = 362881 is not a square number

10! + 1 = 3628800 + 1 = 3628801 is not a square number

So the unsolved problem is: can we find an integer n larger than 7, such as n! + 1 = x^2 ?

Fun with num3ers