Gradient of a higher Lie derivative of a composite function

Question

Although there are some formulations of the gradient of the $k$-th Lie derivative of a function composition $g\circ h$ with respect to $\bf x$ on the vector field $\bf f$ in the literature, I am trying to prove the following pattern for any $k$.

$$ \nabla L_{\bf f}^k(g\circ h) \stackrel{?}{=}\sum_{i=0}^{k}{k \choose i}L_{\bf f}^{k-i}(g'\circ h) \nabla L_{\bf f}^ih $$

This pattern is the second and more reasonable one I came up with after being able to write down the following equalities using brute force.

\begin{align} \frac{\partial L_\mathbf{f}(g\circ h)}{\partial x_i}=& \frac{\partial g}{\partial h}\frac{\partial L_\mathbf{f}h}{\partial x_i}+L_\mathbf{f}(\frac{\partial g}{\partial h})\frac{\partial h}{\partial x_i}\\ \frac{\partial L_\mathbf{f}^2(g\circ h)}{\partial x_i}=& \frac{\partial g}{\partial h}\frac{\partial L_\mathbf{f}^2h}{\partial x_i}+L_\mathbf{f}^2(\frac{\partial g}{\partial h})\frac{\partial h}{\partial x_i}+2L_\mathbf{f}(\frac{\partial g}{\partial h})\frac{\partial L_\mathbf{f}h}{\partial x_i}\\ \frac{\partial L_\mathbf{f}^3(g\circ h)}{\partial x_i}=& \frac{\partial g}{\partial h}\frac{\partial L_\mathbf{f}^3h}{\partial x_i}+L_\mathbf{f}^3(\frac{\partial g}{\partial h})\frac{\partial h}{\partial x_i}+3L_\mathbf{f}^2(\frac{\partial g}{\partial h})\frac{\partial L_\mathbf{f}h}{\partial x_i}+3L_\mathbf{f}(\frac{\partial g}{\partial h})\frac{\partial L_\mathbf{f}^2h}{\partial x_i}&& \end{align}

I would appreciate any help to show that the conjecture above is true/false.

Notation

1. $\nabla$ is the gradient operator such that \begin{align} \nabla \phi = \left( \begin{matrix} \frac{\partial \phi}{\partial x_1}& \frac{\partial \phi}{\partial x_2}& \cdots& \frac{\partial \phi}{\partial x_n} \end{matrix} \right)^\top,\nonumber \end{align} for some function $\phi:\mathbb{ R}\rightarrow\mathbb{ R}$ with respect to a coordinate system $(x_1, x_2, \dots, x_n)$.

2. The Lie derivative of a function $h:\mathbb{ R}^n\rightarrow \mathbb{ R}$ with respect to a vector field $f:\mathbb{ R}^n\rightarrow \mathbb{ R}^n$ is \begin{align} L_{\bf f}(h) := \nabla h \cdot f\,. \nonumber \end{align} One can calculate the $k$-th Lie derivative of $h$ recursively via \begin{align} L^k_{\bf f}(h)=L_{\bf f}L^{k-1}_{\bf f}(h)\,.\nonumber \end{align}

3. We use $g\circ h$ to denote the composition of the functions $h:\mathbb{ R}^n\rightarrow \mathbb{ R}$ and $g:\mathbb{ R}\rightarrow \mathbb{ R}$. We assume that $g$, $h$, and the vector-valued function $f$ are at least $k$ times differentiable. We use $g'$ or $g^{(1)}$ to denote the first derivative of $g$ with respect to $h$. The $j$-th derivative of $g$ is $g^{(j)}$ and $g^{(0)}=g$. The expression $g'\circ h$ has the same meaning as the expression $\partial g/\partial h$.

Answer 1

At each step, except the initial state, we have a sum of products of $2$ factors. Applying $L_\mathbf{f}$ involves taking the gradient, which, by the product rule, produces $2$ sums of $2$ factors for each $1$ we had. In one of them, we apply an extra $L_\mathbf{f}$ in the left factor, in the other one, in the right factor. So, the number of times we end up with $L_\mathbf{f}^i$ in the right factor of the expansion of $\nabla L_\mathbf{f}^k(g\circ h)$ is the number of $i$-subsets of $\{1,\dots,k\}$, i.e. $\binom{k}{i}$.