Proposition 18.4.4 (Power Rule). Let $E$ be a topological vector space, $\sigma \subset B(E)$ be an upward-directed family that includes bounded sets contained in finite-dimensional subspaces, $F$ be a Hausdorff locally convex space, and
be symmetric. For any $x \in E$ and $1 \le k \le n$, let $x^{(k)}$ denote the tuple of $x$ repeated $k$ times, then:
The mapping $f: E \to F \quad x \mapsto T(x^{(n)})$ is infinitely $\sigma$-differentiable on $E$.
For each $1 \le k \le n$ and $x, h \in E$,
\[Df(x)(h_{1}, \cdots, h_{k}) = \frac{n!}{(n-k)!}T(x^{(n-k)}, h_{1}, \cdots, h_{k})\]In particular, $D^{k}f = n! \cdot T$.
For each $k > n$ and $x \in E$, $Df(x) = 0$.
Proof. Suppose inductively that (2) holds for $0 \le k \le n$. Let $G = B^{k}_{\sigma}(E; F)$, then $D^{k}_{\sigma} f \in B^{n-k}_{\sigma}(E; G)$ under the identification $B^{n}_{\sigma}(E; F) = B^{n-k}_{\sigma}(E; B^{k}_{\sigma}(E; F))$ in Proposition 8.11.3. By Theorem 18.4.3, $D^{k}_{\sigma} f$ is also symmetric, so using the Binomial formula,
For each $k \ge 2$, let $A \in \sigma$ and $U \in \cn_{F}(0)$, then since $D^{k}_{\sigma} f \in B^{n-k}_{\sigma}(E; F)$, there exists $t > 0$ such that
for all $s \in (0, t)$. Hence $r \in \mathcal{R}_{\sigma}(E; G)$, and
by the inductive hypothesis.
(3): Since $D^{n}_{\sigma} f$ is constant, $D^{k}_{\sigma} f = 0$ for all $k > n$.$\square$