I came across this funny problem and I wanted to share a very nice solution to explain the behavior of such fractions in any numerical basis (the fraction in the title is in base 10, obviously :). So here it is; beginning with the following definition:
\[
S_n(a) = 1 + 2a + ... + na^{n-1} = \sum_{k=1}^n k a^{k-1} = \frac{ n a^{n+1} - (n+1) a^n + 1 }{ (a-1)^2 }
\]
Given this, it's easy to see that
\[
\frac{ [\ (b-1), (b-2), ...\ , 1\ ]_b }{ [\ 1, 2, ...\ , (b-1)\ ]_b } = \frac{ S_{b-1}(b) }{ b^{b-2} S_{b-1}(1/b) } = \frac{ 1 + (b-2)\ b^b }{ b^b - b(b-1) -1 }
\]
In particular, the fraction in the title equals (with $b=10$):
\[
\frac{ 1 + 8.10^{10} }{ 10^{10} - 91 } \approx \frac{ 1 + 8.10^{10} }{ 10^{10} } \approx 8
\]
