Product rule for certain vector valued functions, or possibly a higher order Chain Rule I'm having trouble proving a composite of $\mathcal{C}^k$ functions is $\mathcal{C}^k$ without use of partial differentiation. To set up my question, consider the following $\mathcal{C}^k$ functions defined on open sets:
$\begin{matrix} U & \xrightarrow{f} & V & \xrightarrow{g} & \mathbb{R}^p, \\ \cap & & \cap & & \\ \mathbb{R}^n & & \mathbb{R}^m\end{matrix}$
To prove the composite $g \circ f:U \to \mathbb{R}^p$ is $\mathcal{C}^k$, it is usually suggested to start by considering a single application of chain rule:
$D(g \circ f)(p) = Dg(f(p)) \circ Df(p). \tag{1}$
I would like to then apply the chain rule again, but the operation $\circ$ above both is and isn't function composition. To explain, it is in the sense that if
If $\,\,\, \alpha = Df(p) \in L(\mathbb{R}^n,\mathbb{R}^m), 
\\ \,\, \beta = Dg(f(p)) \in L(\mathbb{R}^m,\mathbb{R}^p),
\\ \,\, \gamma = D(g \circ f)(p) \in L(\mathbb{R}^n,\mathbb{R}^p) \\ \mathrm{then\,\,} \gamma = \beta \circ \alpha,$
but also isn't, because upon writing
$A = Df: U \to L(\mathbb{R}^n,\mathbb{R}^m) \\ B = (Dg) \circ f : U \to L(\mathbb{R}^m,\mathbb{R}^p) \\ C = D(g \circ f): U \to L(\mathbb{R}^n,\mathbb{R}^p),$ 
$(1)$ becomes $C(p) = B(p) \circ A(p)$, and $C = B \cdot A$, where $\cdot$ is a form of multiplication rather than function composition ($B \circ A$ makes no sense, as their source and target spaces make them incompatible with function composition). Apparently, a form of product rule is needed to find the total derivative of $C = B \cdot A$.
I'm aware that both $A,B$ are Frechet differentiable (presuming $k>1$), but I'm not aware of a product rule regarding total differentiation of anything other than a product of real valued functions. Is there such a rule?
I suspect there is a higher order chain rule as well, yet that which uses a product rule for certain vector valued functions.
 A: If you want to see higher order derivatives in terms of linear maps you have to use multilinear maps (see the post Why is the the $k$-th derivative a symmetric multilinear map?).
I have never seen the higher order chain rule done for multilinear maps. If you use partial derivatives, then what you want is called Faa di Bruno's formula. I am taking $p=1$ (for $p\ge 1$ you apply the formula below to each component of $g$). Hope you are familiar with multi-indeces.
For
every multi-index $\alpha\in\mathbb{N}_{0}^{n}$, with $0<|\alpha|\leq k$,
$$
\frac{\partial^{|\alpha|}}{\partial y^{\alpha}}(g\circ f)(y)=\sum
c_{\alpha,\beta,\gamma,l}\frac{\partial^{|\beta|}g}{\partial x^{\beta}%
}(f(y))\prod_{i=1}^{|\beta|}\frac{\partial^{|\gamma_{i}|}f_{l_{i}}}{\partial
y^{\gamma_{i}}}\left(  y\right)  ,
$$
where $c_{\alpha,\beta,\gamma,l}\in\mathbb{R}$ are constant, the sum is done
over all $\beta\in\mathbb{N}_{0}^{m}$ with $1\leq|\beta|\leq|\alpha|$,
$\gamma=(\gamma_{1},\ldots,\gamma_{|\beta|})$, $\gamma_{i}\in\mathbb{N}
_{0}^{n}$, with $|\gamma_{i}|>0$ and $\sum_{i=1}^{|\beta|}\gamma_{i}=\alpha$,
and $l=(l_{1},\ldots,l_{|\beta|})$, $l_{i}\in\{1,\ldots,m\}$, $i=1, \ldots,
|\beta|$. The proof is by induction on $k$.
You can find it in this paper
https://pdfs.semanticscholar.org/d36c/4938b649ce364aceeb47224f677b3deab0d3.pdf
Having said that, to prove that the composition of functions of class $C^k$ is still $C^k$, I would just do induction on $k$. You prove it for $k=1$ and then assume true for $k-1$. Then you consider the first order partial derivatives of $g\circ f$ which are given in terms of sums products and compositions of partial derivatives of $f$ and of $g$, which are all of (at least) class $C^{k-1}$ and so you can apply the induction hypothesis to get that the first order partial derivatives of $g\circ f$ are of class $C^{k-1}$. In turn, $g\circ f$ is of class $C^k$
