Dimensional countdown
Python, 60 bytes



f=lambda n,k:n&gt;1&lt;k and f(n,k-1)+f(n-1,k)or'987654321\n'[~n:]


Test it on Ideone.

How it works

The k-dimensional countdown from n can be defined with a single base case:


  If n = 1 or k = 1, the output is n || n-1 || ... || 1 || ¶, where || indicates concatenation.


Using the recursive definition from the question, f(n,k) returns f(n,k-1)+f(n-1,k) if n > 1 and k > 1; otherwise it returns the last n + 1 characters from '987654321\n'.