Faà di Bruno's Formula/Proof 2

Proof
For convenience, let:
 * $w_j := D_u^j w$
 * $u_k := D_x^k u$

Then:

Thus:

Analogously, let a corresponding tableau be set up of set partitions as follows:

Thus inspired, let $a_1 a_2 \cdots a_j$ denote a partition of the set $\set {1, 2, \ldots, n-1}$.

Let:

This rule is isomorphic to:

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 $D_x^n w$.

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.