Faà di Bruno's Formula/Example/0/Proof

Theorem
Consider Faà di Bruno's Formula:

When $n = 0$ we have:

Proof
In the summation:
 * $\displaystyle \sum_{j \mathop = 0}^0 D_u^j w \sum_{\substack {\sum_{p \mathop \ge 0} k_p \mathop = j \\ \sum_{p \mathop \ge 0} p k_p \mathop = 0 \\ \forall p \ge 0: k_p \mathop \ge 0} } 0! \prod_{m \mathop = 0}^0 \dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} }$

the only element appearing is for $j = 0$.

Consider the set of $k_p$ such that:
 * $k_0 + k_1 + \cdots = 0$
 * $0 \times k_0 + 1 \times k_1 + \cdots = 0$
 * $k_0, k_1, k_2, \ldots \ge 0$

It is apparent by inspection that:
 * $k_0 = k_1 = \cdots = 0$

When $k_m = 0$:
 * $\dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} } = 1$

by definition of zeroth derivative and factorial of $0$.

Thus any contribution to the summation where $k_m = 0$ can be disregarded.

Thus:
 * $\displaystyle \prod_{m \mathop = 0}^0 \dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} } = 1$

and we are left with: