Google Gemini came up with the answer using a "classic Eulerian technique", multiplying a series by a variable (in this case n) and then subtracting the original series to eliminate terms. After doing this, a geometric series becomes apparent. However, I had to ask the question backwards:

Is there a closed-form to calculate the summation of i.n^i from i = 1 to (n-1)?

The result is:

[n-1,...,2,1,0] (base n) = n^n - n(n^(n-1)-1)/(n - 1)^2

This provides the answer to my original question since the term on the right identifies the distance from n^n, and in particular, the distance from (n^n-1), since:

[n-1,n-1,...,n-1] (base n) = (n^n - 1)

The identity index must be (by symmetry) the same distance from 0, i.e.:

[0,1,2,...,n-1] (base n) = n(n^(n-1)-1)/(n - 1)^2 - 1

Now, what about the permutations of the identity index in between?