10-adic number proof: A^10 has the same n+1 last digits as 1

[tomkoolen]

The question at hand: Let X be a 10-adic number. Let n be a natural number (not 0). Show that A^10 has the same n+1 last digits as 1 if A has the same n last digits as 1 (notation: A =[n]= 1)

My work so far:

(1-X)^10 = (1-X)(1+X+X^2+...+X^10)
A =[n]= 1
1-A =[n]= 0.
I think I can also say that 1+X+X^2+...X^10 =[n]= 1.
But now I don't know how to continue. Could anybody help me out?

Thanks in advance.

 

[tomkoolen]

Of course A = X, my bad!