Let $m \geq 1$ be a fixed integer. There are classical well-known formulas for the sum $\sum\limits_{k=0}^{n} k^{m}$ involving Bernoulli polynomials and/or Stirling numbers. For instance I am particularly interested in the formula $$ \sum\limits_{k=0}^{n} k^{m} = \sum\limits_{j \geq 0} \dfrac{S(m,j)}{j+1} \left(\sum\limits_{k \geq 0} s(j+1, k) (n+1)^{k} \right) $$ where $S(-,-)$ and $s(-,-)$ denote the Stirling numbers of the second and first kinds respectively (I mean signed Stirling numbers of the first kind). (See Concrete Mathematics by R.L. Graham, D. Knuth and O. Patashnik for a proof).
I wonder if there is such a formula for the $r$-fold sum of powers of integers in terms of Stirling numbers of both kinds. Recall that the $r$-fold sum is defined recursively as $$ {\sum}^{1} n^m = \sum\limits_{k=0}^{n} k^{m} = 1^m + 2^m + \ldots + n^m, $$ $$ {\sum}^{r} n^m = {\sum}^{r-1} 1^m + {\sum}^{r-1} 2^m + \ldots + {\sum}^{r-1} n^m,\quad r \geq 2 $$
It seems to be complicated to obtain a similar formula for this $r$-fold sum using this recursive definition. Any result or reference is appreciated.