$ n^2+n=(n+1)\times n=1 \times n^2+n \times 1 \Rightarrow k(n) \le n+1$ Assume that $ k(n) \le n$ then $ \exists m| m=a_0 \times 1+ a_1 \times n+\cdots+a_c \times n^c= b_0 \times 1+ b_1 \times n+\cdots+b_c \times n^c+\cdots+ b_d \times n^d$ and $ a_1+a_2+\cdots+a_c= b_1+b_2+\cdots+b_c+\cdots+b_d=k(n)$ If $ a_i=n \Rightarrow m=n \times n^i=n^{i+1}$cannot be written as a different sum of $ n$ terms of $ S_n$ then $ a_i,b_i < n$ then $ 0= (b_0-a_0) \times 1+ (b_1-a_1) \times n+\cdots+(b_c-a_c) \times n^c+b_{c+1} \times n^{c+1}+\cdots+ b_d \times n^d$ Pass all the negative terms to the other side and noticing that in the LHS each coefficient $ a_q-b_q$ is less than $ n$ then each term $ (a_q-b_q) \times n^q$ is less than $ n^q$ then the LHS is less than $ n^{c+1}$ contradiction. Then $ k(n)=n+1$, is it ok?

After you find that k(n) could be n+1 using the n^2+n example, you could just note that any sum that involves equal to or less than n elements of S_n represents the base-n expansion of an integer (there is one counterexample to this, namely if you take n of the same element, but this is trivial to overcome), and as this is clearly UNIQUE, we must have that k(n)>n which implies that k(n) = n+1 as desired. (I'm sorry but I don't know how to use Latex).