I'm trying prove the following identity
$$S(n,k)=\sum_{(\alpha_1,\cdots,\alpha_k)}1^{\alpha_1-1}2^{\alpha_2-1}\cdots k^{\alpha_k-1},$$
where the sum ranges over the $k$-compositions of $n$. I naively attempted to finds a relationship between Stirling numbers and compositions, which led me to the identity
$$S(n,k)=\sum \frac{n!}{b_1!^{m_1}m_1!\cdots b_k!^{m_k}m_k!},$$
where we sum over all the different $k$-compositions $(b_1,\cdots,b_k)$ of $n$ where $b_1\leq \cdots \leq b_k$ and $m_i$ is the number of times $b_i$ appears in the $k$-composition. This, as far as I can tell, brought me no closer to solving the original problem. Any suggestions?
Here is a tentative demonstration: Starting from the generating function $$\sum_{n\ge0}S(n+k,k)x^n=\prod_{r=1}^k\frac{1}{1-rx }$$ $$=\prod_{r=1}^k\sum_{i\ge0}r^ix^i$$ $$=\sum_{n\ge0}(\sum_{(a_i)\in A_n}\prod_{r=1}^kr^{a_r})x^n$$ where $A_n=\{(a_i);0\ge a_i\ge k;\sum_{i=1}^ka_i=n\}$ then $$S(n,k)=\sum_{(a_i)\in A_{n-k}}\prod_{r=1}^kr^{a_r}$$ let $B_n=\{(b_i);1\ge b_i\ge k;\sum_{i=1}^kb_i=n\}$ and $C_n=\{(b_i-1);(b_i)\in B_n \}$ then it is not difficult to see that $C_n=A_{n-k}$ then $$S(n,k)=\sum_{(b_i)\in B_n}\prod_{r=1}^kr^{b_r-1}$$ which is what we wanted to show.