The radix complement is most easily obtained by adding 1 to the diminished radix complement, which is (b

^{n}-1)-y. Since b

^{n}-1 is the digit b-1 repeated n times because

b

^{n}-1 = b

^{n}-1

^{n}= (b-1)(b

^{n-1}+b

^{n-2}...+b+1) = (b-1)b

^{n-1}+ (b-1)b

^{n-2}+...+b-1

I'm trying to understand the third step in this pattern. I understand that an intermediary step would be (b-1)

^{n}but why does this translate into (b-1)(b

^{n-1}+b

^{n-2}...+b+1)?

Thanks,

deersfeet