Let $S_{n}=\{1,n,n^{2},n^{3}, \cdots \}$, where $n$ is an integer greater than $1$. Find the smallest number $k=k(n)$ such that there is a number which may be expressed as a sum of $k$ (possibly repeated) elements in $S_{n}$ in more than one way. (Rearrangements are considered the same.)
1. Consider the set \( S_n = \{1, n, n^2, n^3, \ldots\} \) where \( n \) is an integer greater than 1. We need to find the smallest number \( k = k(n) \) such that there is a number which can be expressed as a sum of \( k \) (possibly repeated) elements in \( S_n \) in more than one way.

2. Let's start by considering the sum \( n^2 + n \). This can be written as:
   \[
   n^2 + n = n \cdot n + n \cdot 1 = 1 \cdot n^2 + n \cdot 1
   \]
   This shows that \( n^2 + n \) can be expressed as a sum of \( n + 1 \) elements in \( S_n \) in more than one way. Therefore, \( k(n) \leq n + 1 \).

3. Now, assume for contradiction that \( k(n) \leq n \). This means there exists a number \( m \) that can be expressed as a sum of \( k \) elements in \( S_n \) in two different ways:
   \[
   m = a_0 \cdot 1 + a_1 \cdot n + \cdots + a_c \cdot n^c = b_0 \cdot 1 + b_1 \cdot n + \cdots + b_c \cdot n^c + \cdots + b_d \cdot n^d
   \]
   where \( a_1 + a_2 + \cdots + a_c = b_1 + b_2 + \cdots + b_c + \cdots + b_d = k(n) \).

4. If \( a_i = n \), then \( m = n \cdot n^i = n^{i+1} \), which cannot be written as a different sum of \( n \) terms of \( S_n \). Therefore, \( a_i, b_i < n \).

5. Subtract the two expressions for \( m \):
   \[
   0 = (b_0 - a_0) \cdot 1 + (b_1 - a_1) \cdot n + \cdots + (b_c - a_c) \cdot n^c + b_{c+1} \cdot n^{c+1} + \cdots + b_d \cdot n^d
   \]
   Move all negative terms to the other side:
   \[
   (a_0 - b_0) \cdot 1 + (a_1 - b_1) \cdot n + \cdots + (a_c - b_c) \cdot n^c = b_{c+1} \cdot n^{c+1} + \cdots + b_d \cdot n^d
   \]
   Since each coefficient \( a_q - b_q \) is less than \( n \), each term \( (a_q - b_q) \cdot n^q \) is less than \( n^{q+1} \). Therefore, the left-hand side (LHS) is less than \( n^{c+1} \), which is a contradiction because the right-hand side (RHS) is at least \( n^{c+1} \).

6. Hence, \( k(n) = n + 1 \).

The final answer is \( \boxed{ k(n) = n + 1 } \).