1. y = x^5 - x = (x - 1)*x*(x + 1)*(x^2 + 1)
Therefore y = 0 (mod 5) and y = 0 (mod 2). It means x^5 and x are ended with the same digit.

2. (x + 1)^5 = (x^5 + 5*x^4 + 10*x^3 + 10*x^2 + 5*x + 1) =
5*x*(1 + x)*(1 + x + x^2) + (x^5 + 1)
Easy to show 5*x*(1 + x)*(1 + x + x^2) = 0 (mod 10)
Therefore the last digit of (x + 1)^5 is the last digit of x^5 plus 1. Start with x = 0 we can get all the last digits of 0, 1, 2, ... 9 are the same 0, 1, 2,...9

May be exist also another interesting proofs?

I am thinking now about two small questions:
1. Why after one, five is the first digit with this property?
2. Why other one, only 5 and 6 hold the property: the last digit of x^n is the same x with any n?

Best regards,
Bui Quang Tuan

Because 10 = 2*5 and 2 is too small.

>2. Why other one, only 5 and 6 hold the property: the last
>digit of x^n is the same x with any n?

Base
6	3² = 13 4² = 24
10	5² = 25 6² = 36
12	4² = 14 9² = 69
14	7² = 37 8² = 48
15	6² = 26 A² = 6A
18	9² = 49 A² = 5A
21	7² = 27