The previous post gives the 2500th digit reading from left to right, but the puzzle requires reading from right to left.

How many zeroes are at the end of 10000! ?

Multiples of 5 = 2000

+

Multiplies of 25 = 400 (each of which add one more 5)

+

Multiples of 125 = 80

+

Multiples of 625 = 16

+

Multiples of 3125 = 3,

so the last 2499 digits are zeroes.

10! = 10*9*8*7*6*5*4*3*2*1 = 3628**8**00,

so I think the final answer is 8^1000 (mod 10).

= 4^500 (mod 10) (because 8*8 = 6**4**)

= 4^100 (mod 10) (because 4*4*4*4*4 mod 10 = 4)

= 4^20 (mod 10)

= 4^4 (mod 10)

= 6

Unless, of course, I've made a mistake, which has been known to happen