Let $n,k$ be constant integers. What is as tight upper bound on $S$: $$S_{n,k}=\sum_{i_1=1}^{n-k}\sum_{i_2=1}^{n-i_1}\dots\sum_{i_k=1}^{n-i_{k-1}} \frac{1}{i_1 i_2\dots i_k}$$
It is easy to show that $S\le H(n)^k$ in which $H(n)$ is the harmonic sum. Because every term in $S_{n,k}$ also appears bellow: $$S_{n,k}<\sum_{i_1=1}^n\frac{1}{i_1}\sum_{i_2=1}^n\frac{1}{i_2}\dots\sum_{i_k=1}^n \frac{1}{i_k} = H(n)^k$$ However, my question is, can a tigher bound be achieved since many terms are not present in $S$? For instance, is thera a multiplicative factor like $\alpha(n,k)<1$ such that $S_{n,k}\le \alpha(n,k) H(n)^k$?
One possible way to estimate or bound $S_{n,k}$ is the following recursive relationship: $$S_{n,k}=\sum_{i=1}^{n-k} \frac{1}{i} S_{n-i,k-1}$$ and $$S_{n,1}=H(n-1)$$ Does this lead to any bound?