Remainder Theorems

For all prime numbers p, when (p-2)! Is divided by p, the remainder is 1

323 = 17 * 19

By Wilson's theorem we can say that.

16! + 1 is divisible by 17.
==> (16! +1)+85 is also divisible by 17.(Since 85 is divisible by 17)

18!+1 is divisible by 19.

18! + 1 = 0(mod 19)
18*17*16! + 1 = 0 (mod 19)
==> (-2)*(-1)*16! +1 = 0(mod 19) (Since 18= -1(mod19) and 17 = -2(mod 19))

==> 2*16! +1 = 0(mod 19)

This can only possible when 16! = 9(mod 19)
==> 16! + 10 = 0(mod 19)

==>16! + 86 = 0(mod 19)

Therefore the remainder when 16!+86 is divided by 323 is 0.

wkt $$(n+1)!-n! = (n+1)\times n!-n!$$ = $$n\times n!+1\times n!-n! =n\times n!$$
Hence $$1*1! + 2*2! + 3*3! + . . . + 63*63! +64*64! = (2!-1!)+(3!-2!)+...+(65!-64!) = 65!-1!$$
Or we can expand the above expression as 
$$(2-1)*1!+(3-1)*2!+(4-1)*3!+...+(65-1)*64!$$ = $$2!-1!+3!-2!+4!-3!+...+65!-64!$$ = $$65!-1!$$
Wilson's theorem on prime numbers suggests that Reminder of $$\frac{(p-2)!}{p}$$ is 1 where p is prime
Hence, when 65! is divided by 67, it leaves a remainder of 1.
So, the reminder of $$\frac{65!-1}{67}$$ is 1-1 = 0.