For convenience, let: :$w_j := D_u^j w$ :$u_k := D_x^k u$ Then: {{begin-eqn}} {{eqn | l = D_x \left({w_j}\right)       | r = w_{j + 1} u_1       | c = Derivative of Composite Function }} {{eqn | l = D_x \left({u_k}\right)       | r = u_{k + 1}       | c = {{Defof|Higher Derivative}} }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = D_x^1 w       | r = w_1 u_1       | c = Derivative of Composite Function }} {{eqn | l = D_x^2 w       | r = \left({w_2 u_1 u_1 + w_1 u_2}\right)       | c = Derivative of Composite Function: Second Derivative }} {{eqn | l = D_x^3 w       | r = \left({\left({w_3 u_1 u_1 u_1 + w_2 u_2 u_1 + w_2 u_1 u_2}\right) + \left({w_2 u_1 u_2 + w_1 u_3}\right)}\right)       | c = Derivative of Composite Function: Third Derivative or directly }} {{end-eqn}} Analogously, let a corresponding tableau be set up of set partitions as follows: {{begin-eqn}} {{eqn | l = \mathcal D^1       | r = \set 1       | c =  }} {{eqn | l = \mathcal D^2       | r = \set {\set 2 \mid \set 1} + \set {2, 1}       | c =  }} {{eqn | l = \mathcal D^3       | r = \set {\set 3 \mid \set 2 \mid \set 1} + \set {\set {3, 2} \mid \set 1} + \set {\set 2 \mid \set {3, 1} } + \set {\set 3 \mid \set {2, 1} } + \set {3, 2, 1}       | c =  }} {{end-eqn}} Thus inspired, let $a_1 a_2 \cdots a_j$ denote a partition of the set $\set {1, 2, \ldots, n}$. Let: {{begin-eqn}} {{eqn | l = \mathcal D a_1 a_2 \cdots a_j       | o = :=       | r = \set n a_1 a_2 \cdots a_j       | c =  }} {{eqn | ro= +       | o =        | r = \left({a_1 \cup \set n}\right) a_2 \cdots a_j        | c =  }} {{eqn | ro= +       | o =        | r = a_1 \left({a_2 \cup \set n}\right) \cdots a_j        | c =  }} {{eqn | ro= +       | o =        | r = \cdots       | c =  }} {{eqn | ro= +       | o =        | r = a_1 a_2 \cdots \left({a_j \cup \set n}\right)       | c =  }} {{end-eqn}} This rule is isomorphic to: {{begin-eqn}} {{eqn | l = D_x \left({w_j u_{r_1} u_{r_2} \ldots u_{r_j} }\right)       | r = w_{j + 1} u_1 u_{r_1} u_{r_2} \ldots u_{r_j}       | c =  }} {{eqn | ro= +       | o =        | r = w_j u_1 u_{r_1 + 1} u_{r_2} \ldots u_{r_j}       | c =  }} {{eqn | ro= +       | o =        | r = w_j u_1 u_{r_1} u_{r_2 + 1} \ldots u_{r_j}       | c =  }} {{eqn | ro= +       | o =        | r = \cdots       | c =  }} {{eqn | ro= +       | o =        | r = w_j u_1 u_{r_1} u_{r_2} \ldots u_{r_j + 1}       | c =  }} {{end-eqn}} where the term $w_j u_{r_1} u_{r_2} \ldots u_{r_j}$ corresponds to a partition $a_1 a_2 \cdots a_j$. Thus there exists a bijection from $\mathcal D^n$ to $D_x^n w$. {{explain|Link this to Number of Set Partitions by Number of Components}} Thus, collecting like terms in $D_x^n w$, we obtain a sum of terms: :$c \left({k_1, k_2, \ldots}\right) w_j u_i^{k_1} u_2^{k_2} \ldots$ where: :$j = k_1 + k_2 + \cdots$ and: :$n = k_1 + 2 k_2 + \cdots$ and where: :$c \left({k_1, k_2, \ldots}\right)$ is the number of partitions of $\set {1, 2, \ldots, n}$ into $j$ subsets where there are $k_t$ subsets with $t$ elements. Consider an array of $k_t$ boxes of capacity $t$. The number of ways to put $n$ different elements into these boxes is the multinomial coefficient: :$\dbinom n {1, 1, \ldots, 1, 2, 2, \ldots, 2, 3, 3, \ldots, 3, 4, \ldots} = \dfrac {n!} {\left({1!}\right)^{k_1} \left({2!}\right)^{k_2} \left({3!}\right)^{k_3} \cdots}$ It remains to divide by $k_1! \, k_2! \, k_3! \ldots$ corresponding to the number of ways each group of $k_t$ can be permuted Hence: :$c \left({k_1, k_2, \ldots}\right) = \dfrac {n!} {k_1! \left({1!}\right)^{k_1} \, k_2! \left({2!}\right)^{k_2} k_3! \left({3!}\right)^{k_3} \cdots}$ and the result follows. {{qed}}