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I don't know if this serves counts as a combinatorial interpretation, but the identity can be seen as follows: Firstly it's equivalent to $$m\sum_{k=0}^m a_k c_{m-k} = (n+1)\sum_{k=0}^mka_kc_{m-k}\hspace{40pt}(1)$$ If we write $f(x) = \sum_{k=0}^\infty a_kx^k$ then $\sum_{k=0}^m a_k c_{m-k}$ is the $m$-th coefficient of $f^{n+1}(x)$ and the LHS resp. RHS of $(1)$ is the $(m-1)$-th coefficient of $$(f^{n+1})'(x)=(n+1)f^n(x)f'(x)\hspace{50pt}(2)$$ But $f'(x)=\sum_{k=0}^\infty (k+1)a_{k+1}x^k$. Hence just the $(m-1)$-th coefficient of the RHS of $(2)$ is $$(n+1)\sum_{k=0}^{m-1}(k+1)a_{k+1}c_{m-1-k}=(n+1)\sum_{k=1}^m ka_kc_{m-k}$$ which equals the LHS resp. RHS of $(1)$. $(f^{n+1})'=(n+1)f^nf'.\hspace{50pt}$$1 I don't know if this serves as a combinatorial interpretation, but the identity can be seen as follows: Firstly it's equivalent to$$m\sum_{k=0}^m a_k c_{m-k} = (n+1)\sum_{k=0}^mka_kc_{m-k}\hspace{40pt}(1)$$If we write f(x) = \sum_{k=0}^\infty a_kx^k then \sum_{k=0}^m a_k c_{m-k} is the m-th coefficient of f^{n+1}(x) and the LHS of (1) is the (m-1)-th coefficient of$$(f^{n+1})'(x)=(n+1)f^n(x)f'(x)\hspace{50pt}(2)$$But f'(x)=\sum_{k=0}^\infty (k+1)a_{k+1}x^k. Hence the (m-1)-th coefficient of the RHS of (2) is$$(n+1)\sum_{k=0}^{m-1}(k+1)a_{k+1}c_{m-1-k}=(n+1)\sum_{k=1}^m ka_kc_{m-k}$$which equals the RHS of$(1)\$.