Calculate a multiple sum of inverse integers. The question is to calculate a following sum:
\begin{equation}
{\mathcal S}_p(n) :=\sum\limits_{1\le j_1 < j_2 < \dots <j_p \le n-1} \prod\limits_{q=1}^p \frac{1}{n-j_q}
\end{equation}
for $p=1,2,..$ and $n\ge 1$.
From purely combinatorial reasoning we have:
\begin{eqnarray}
{\mathcal S}_1(n) &=& H_{n-1} \\
{\mathcal S}_2(n) &=& \frac{1}{2!} \left(H_{n-1}^2 - H^{(2)}_{n-1} \right) \\
{\mathcal S}_3(n) &=& \frac{1}{3!} \left(H_{n-1}^3 - 3 H_{n-1} H_{n-1}^{(2)} + 2 H_{n-1}^{(3)}\right) \\
{\mathcal S}_4(n) &=& \frac{1}{4!} \left(H^4_{n-1} - 6 H_{n-1}^2 H_{n-1}^{(2)} + 8 H_{n-1} H_{n-1}^{(3)} + 3 H_{n-1}^{(2)} H_{n-1}^{(2)} - 6 H_{n-1}^{(4)}\right) \\
{\mathcal S}_5(n) &=& \frac{1}{5!} \left(H_{n-1}^5 - 10 H_{n-1}^3 H_{n-1}^{(2)} + 20 H_{n-1}^2 H_{n-1}^{(3)} + 15 H_{n-1} ((H_{n-1}^{(2)})^2 - 2 H_{n-1}^{(4)}) - 20 H_{n-1}^{(2)} H_{n-1}^{(3)}  + 24 H_{n-1}^{(5)}\right) \\
{\mathcal S}_6(n) &=& \frac{1}{6!} \left(H_{n-1}^6 - 15 H_{n-1}^4 H_{n-1}^{(2)} + 40 H_{n-1}^3 H_{n-1}^{(3)} + 45 H_{n-1}^2 ((H_{n-1}^{(2)})^2 - 2 H_{n-1}^{(4)}) - 
 24 H_{n-1} (5 H_{n-1}^{(2)} H_{n-1}^{(3)} - 6 H_{n-1}^{(5)}) + 5 (-3 (H_{n-1}^{(2)})^3 + 18 H_{n-1}^{(2)} H_{n-1}^{(4)} + 8 ((H_{n-1}^{(3)})^2 - 3 H_{n-1}^{(6)}) \right)
\end{eqnarray}
where $H_{n-1}^{(r)} := \sum\limits_{j=1}^{n-1} 1/j^r$ is the generalised Harmonic number.
Is it possible to find the result for generic $p\ge 1$?
 A: Expanding the right hand side of the identity given by achille hui  we get ``a compact'' expression for the sum:
\begin{eqnarray}
{\mathcal S}_p(n) &=& \sum\limits_{m=1}^p \frac{(-1)^{-m+p} }{m!} \sum\limits_{p_1+p_2+\dots+p_m=p} \prod\limits_{q=1}^m \frac{H_{n-1}^{(p_q)}}{p_q} \\
&=& \frac{1}{1!} \frac{(-1)^{p-1}}{p} H^{(p)}_{n-1} + \frac{(-1)^{p-2}}{2!} \sum\limits_{p_1=1}^{p-1} \frac{1}{p_1 (p-p_1)} H_{n-1}^{(p_1)} H_{n-1}^{(p-p_1)} + \dots + \frac{1}{p!} H_{n-1}^p
\end{eqnarray}
A: Suppose we seek to evaluate
$$S_n(m) = \sum_{1\le j_1 \lt j_2 \lt\cdots\lt j_n \le m-1}
\prod_{q=1}^n \frac{1}{m-j_q}.$$
By way of enrichment let me point  out that we can express this sum in
terms  of the  cycle  index  $Z(P_n)$ of  the  unlabeled set  operator
$\mathfrak{P}_{=n}$ by applying the Polya Enumeration Theorem.

Recall  the recurrence by Lovasz  for the cycle  index $Z(P_n)$ of
the set operator $\mathfrak{P}_{=n}$ on $n$ slots, which is
$$Z(P_n) = \frac{1}{n} \sum_{l=1}^n (-1)^{l-1} a_l Z(P_{n-l})
\quad\text{where}\quad
Z(P_0) = 1.$$
This recurrence lets us calculate the cycle index $Z(P_n)$ very easily.

The sum is then given by
$$Z(P_n)(Q_1+Q_2+\cdots+Q_{m-1})$$
evaluated at $Q_k = \frac{1}{m-k}.$
The Polya enumeration rule says to substitute as follows:
$$a_l = Q_1^l + Q_2^l + \cdots + Q_{m-1}^l$$
in other words
$$a_l = H_{m-1}^{(l)}$$
for a final answer of 
$${\large
\bbox[5px,border:2px solid #00A000]{ \left.Z(P_n)\right|_{a_l = H_{m-1}^{(l)}}}}$$
For example we have
$$Z(P_5) = \frac{1}{5!}
\left({a_{{1}}}^{5}-10\,a_{{2}}{a_{{1}}}^{3}+20\,a_{{3}}{a_{{1}}}^{2}
+15\,a_{{1}}{a_{{2}}}^{2}-30\,a_{{4}}a_{{1}}
-20\,a_{{2}}a_{{3}}+24\,a_{{5}}\right)$$
and the substitution should now be clear.
