The proof proceeds by induction on $n$. For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } n! \prod_{m \mathop = 1}^n \dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} }$ The case for $n = 0$ is: {{:Faà di Bruno's Formula/Example/0}} which is consistent with the definition of the zeroth derivative. The case for $n = 1$ is: {{:Faà di Bruno's Formula/Example/1}} which is consistent with Derivative of Composite Function. === Basis for the Induction === $P \left({2}\right)$ is the case: {{:Faà di Bruno's Formula/Example/2}} This is consistent with Derivative of Composite Function: Second Derivative. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 2$, then it logically follows that $P \left({r + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle D_x^r w = \sum_{j \mathop = 0}^r D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = r \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } r! \prod_{m \mathop = 1}^r \dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} }$ from which it is to be shown that: :$\displaystyle D_x^{r + 1} w = \sum_{j \mathop = 0}^{r + 1} D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = r \mathop + 1 \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } \left({r + 1}\right)! \prod_{m \mathop = 1}^{r + 1} \dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} }$ === Induction Step === This is the induction step: Note that when $k_m = 0$: :$\dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} } = 1$ which shows that any contribution to the summation where $k_m = 0$ can be disregarded. Let $j = 0$. Consider the set of $k_p$ such that: :$k_1 + k_2 + \cdots = 0$ :$1 \times k_1 + 2 k_2 + \cdots = r$ :$k_1, k_2, \ldots \ge 0$ It is apparent by inspection that, for all $r > 0$, no set of $k_p$ can fulfil these conditions. Therefore when $j = 0$ the summation is vacuous. Also note that from: :$1 \times k_1 + 2 k_2 + \cdots = r$ it follows that: :$k_{r + 1} = k_{r + 2} = \cdots = 0$ Thus, while there are only finitely many $k$'s, their upper limit need not be explicitly considered. Let $c \left({r, j, k_1, k_2, \ldots}\right)$ be the coefficient of $D_u^j w$ in $D_x^r w$. We establish some lemmata: Lemma 1: {{:Faà di Bruno's Formula/Lemma 1}} Lemma 2: {{:Faà di Bruno's Formula/Lemma 2}} {{finish}} By differentiating {{WRT|Differentiation}} $x$: {{begin-eqn}} {{eqn | l = c \left({r + 1, j, k_1, k_2, \ldots}\right)       | r = c \left({r, j - 1, k_1 - 1, k_2, \ldots}\right)       | c =  }} {{eqn | o =       | ro= +       | r = \left({k_1 + 1}\right) c \left({r, j, k_1 + 1, k_2 - 1, k_3, \ldots}\right)       | c =  }} {{eqn | o =       | ro= +       | r = \left({k_2 + 1}\right) c \left({r, j, k_1, k_2 + 1, k_3 - 1, k_4, \ldots}\right)       | c =  }} {{eqn | o =       | ro= +       | r = \cdots       | c =  }} {{end-eqn}} The equations: :$k_1 + k_2 + \cdots = j$ and: :$k_1 + 2 k_2 + \cdots = r$ are preserved by this induction step. Thus it is possible to factor out: :$\dfrac {r!} {k_1! \left({1!}\right)^{k_1} \cdots k_r! \left({r!}\right)^{k_r} }$ from each term on the {{RHS}} of the equation for $c \left({r + 1, j, k_1, k_2, \ldots}\right)$. Thus we are left with: :$k_1 + 2 k_2 + 3 k_3 + \cdots = r + 1$ So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 0}: D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } n! \prod_{m \mathop = 1}^n \dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} }$