Question 7 Calculate $\omega_{\mathrm{D}}=\sum_{\substack{\mathrm{A} \subset \mathrm{S} \\ \mathrm{A} \neq \varnothing}}\left(\prod_{\mathrm{a} \in \mathrm{A}} a\right)$.
Let $n \in A \subset S$ and $|A| \geqslant 2$, define $f(A)=A - \{n\}$, then $f(A) \neq A$ and $f(A) \subset \{1,2, \cdots, n-1\} - S$. Notice that
\[
\prod_{a \in A} a + \prod_{a \in f(A)} a = n \prod_{a \in f(A)} a + \prod_{a \in f(A)} a = (n+1) \prod_{a \in f(A)} a
\]
only the singleton subsets $\{n\}$ of $S$ do not have a pair, so
\[
\begin{array}{l}
- \left(\prod_{a \in A} a\right) + n = (n+1) \omega_{0-1} + n, \\
\end{array}
\]
which means $\omega_{n} + 1 = (n+1) (\omega_{n-1} + 1)$.
Let $\varepsilon_{n} = \omega_{n} + 1$, then we have the recursive relation
\[
\left\{\begin{array}{l}
\varepsilon_{n} = (n+1) \varepsilon_{n-1}, \\
\varepsilon_{1} = 2
\end{array}\right.
\]
Thus, $\varepsilon_{n} = (n+1) \varepsilon_{n-1} = (n+1) n \varepsilon_{n-2}$
\[
= (n+1) n \cdots 3 \varepsilon_{1} = (n+1)!.
\]
Therefore, $\omega_{n} = (n+1)! - 1$.