Consider the set $S=\{1,2,...,n\}$. For every $k\in S$, define $S_{k}=\{X \subseteq S, \ k \notin X, X\neq \emptyset\}$. Determine the value of the sum \[S_{k}^{*}=\sum_{\{i_{1},i_{2},...,i_{r}\}\in S_{k}}\frac{1}{i_{1}\cdot i_{2}\cdot...\cdot i_{r}}\] [hide]in fact, this problem was taken from an austrian-polish[/hide]
1. Consider the set \( S = \{1, 2, \ldots, n\} \). For every \( k \in S \), define \( S_k = \{X \subseteq S \mid k \notin X, X \neq \emptyset\} \). We need to determine the value of the sum:
   \[
   S_k^* = \sum_{\{i_1, i_2, \ldots, i_r\} \in S_k} \frac{1}{i_1 \cdot i_2 \cdot \ldots \cdot i_r}
   \]

2. First, evaluate \( S_0^* \) (the sum over all non-empty subsets of \( S \)):
   \[
   S_0^* = \sum_{\{i_1, i_2, \ldots, i_r\} \subseteq S} \frac{1}{i_1 \cdot i_2 \cdot \ldots \cdot i_r}
   \]
   For example, when \( n = 3 \):
   \[
   S_0^* = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{1 \cdot 2} + \frac{1}{1 \cdot 3} + \frac{1}{2 \cdot 3} + \frac{1}{1 \cdot 2 \cdot 3}
   \]

3. We can use the following identity to simplify the sum:
   \[
   1 + S_0^* = \left(1 + \frac{1}{1}\right) \left(1 + \frac{1}{2}\right) \cdots \left(1 + \frac{1}{n}\right)
   \]
   This product expands to:
   \[
   \frac{2}{1} \cdot \frac{3}{2} \cdots \frac{n+1}{n} = n + 1
   \]
   Therefore:
   \[
   S_0^* = n
   \]

4. Now, to find \( S_k^* \). If we remove the term \( \frac{1}{k} \) from \( S_0^* \), then for every remaining term with \( k \) in its denominator subset, there is a term with the same subset except without \( k \). So the sum of the latter group is \( k \) times larger than that of the former. But the latter group consists of all the terms without \( k \) in their denominators, therefore its sum is \( S_k^* \). Therefore:
   \[
   S_0^* - \frac{1}{k} = \frac{k+1}{k} S_k^*
   \]

5. Solving for \( S_k^* \):
   \[
   S_k^* = \frac{k S_0^* - 1}{k + 1} = \frac{nk - 1}{k + 1}
   \]
   for \( k \in S \). Otherwise, \( S_k^* = n \).

The final answer is \( \boxed{\frac{nk - 1}{k + 1}} \) for \( k \in S \).