Very easy math! Help!

74 = 2401 = 2400 + 1
(74)k = (2400 + 1)k
= ∑(r=0 to 2400) (kCr)2400r
Any terms higher than 24002 are multiples of 10000.
So ∑(r=0 to 2400)(kCr)2400r can be expressed as 10000N + 2400k + 1 where N is an integer.
It is obvious to see if k is a multiple of 5, the second last term is a multiple of 1000 leaving the whole sum a number with ending digits 001.
Indeed 720n satisfies the required condition for any positive integer n.

n=0 時 , 咩既 n 次都等於 1