Computing a derivative through Lie series 
Consider the $N$-dimensional autonomous system of ODEs
  $$\dot{x}= f(x),$$
  where a locally unique solution $x(t)$, starting from the initial condition $x$, is denoted as $x(t)=\phi(t,x)$. Assume that
$$\Big(\frac{\partial}{\partial{x}}\phi(t,x)\Big)f(x)=f(\phi(t,x))$$
For the system above, assume that $f(x)$ is analytic (that is, its Taylor series converges to $f$ itself). Let the differential operator $L[\xi]$ be defined as 
$$L[\xi]=f(x)\boldsymbol{\cdot}\nabla{\xi}=\sum_{n=1}^{N}f_i(x)\frac{\partial{\xi}}{\partial{x_i}}$$
Show that $\phi(t,x)$ can be expressed as 
$$\phi(t,x)=\sum_{n=0}^{\infty}\frac{t^n}{n!}L^n[x]$$
where $L^n[\xi]$ is the shorthand notation for 
$$L^n[\xi]=\underbrace{L[L[\cdots{L}[\xi]}_{n\text{-times}}\cdots]]$$

Potentially related questions:

How to properly apply the Lie Series
Exponential of a function times derivative
How to derive these Lie Series formulas

I'm stuck on how to approach this problem. Here is all the information that I have gathered so far -
Through this question, the one dimensional situation states that $e^{a\partial}f(x)=f(a+x)$ (we can think of this as a shift operator). 
Inside Ordinary Differential Equations and Dynamical Systems by Teschl, we have the following Lemma (Lemma $6.2$ on page $190$ of the text).
Lemma (Straightening out of vector fields): Suppose $f(x_0)\neq0$. Then, there is a local coordinate transform $y=\varphi(x)$ such that $\dot{x}=f(x)$ is transformed to 
$$\dot{y}=(1,0,...,0)$$
Teschl list a similar problem on page $191$ (problem $6.5$ for one-parameter lie groups) in which he states that 
Hint: The Taylor coefficients are the derivatives which can be obtained by
differentiating the differential equation.
So, I think that I need to apply what was done in this question alongside Lemma 6.2. I will have to consider what a vector field means in this context. I might be able to make the assumption that a vector field is just a linear operator. We are given that 


*

*$\dot{x}= f(x)$ is an autonomous system of ODEs

*$x(t)=\phi(t,x)$

*$\Big(\frac{\partial}{\partial{x}}\phi(t,x)\Big)f(x)=f(\phi(t,x))$

*$L[\xi]=f(x)\boldsymbol{\cdot}\nabla{\xi}=\sum_{n=1}^{N}f_i(x)\frac{\partial{\xi}}{\partial{x_i}}$
and we need to show that
$$\phi(t,x)=\sum_{n=0}^{\infty}\frac{t^n}{n!}L^n[x]$$
I also see that Roger Howe wrote a good introduction to lie theory in these notes (he goes through one-parameter lie groups on pages $604-606$).
This appears to be an extremely difficult problem for someone unfamiliar with lie theory. I am going to see if I can figure out a more direct approach.
 A: For any differentiable function $B:\Bbb R^n\to\Bbb R^n$ we know from chain rule and differential equation that
\begin{align}
\frac{∂}{∂t}B(ϕ(t,x))&=\frac{∂B}{∂x}(ϕ(t,x))\cdot \frac{∂}{∂t}ϕ(t,x)
\\
&=\frac{∂B}{∂x}(ϕ(t,x))\cdot f(ϕ(t,x))
\\
&=\sum_{i=1}^n \frac{∂B}{∂x_i}(ϕ(t,x)) f_i(ϕ(t,x))=L_{ϕ(t,x)}[B].
\end{align}
So along a solution we get $\frac{∂}{∂t}=L_{ϕ(t,x)}$. Now apply this to the translation operator resp. the Taylor expansion
$$
ϕ(t,x)=\exp\left(t\frac{∂}{∂s}\right)ϕ(s,x)\Big|_{s=0}
=\exp\left(tL_{ϕ(s,x)}\right)[ϕ(s,x)]\Big|_{s=0}
=\exp\left(tL_{x}\right)[x]\Big|_{s=0}
$$
The same remains true if you replace the exponential by the exponential series.
A: Here is my interpretation of the first answer.
Suppose we have a differentiable function $\xi:\Bbb R^n\to\Bbb R^n$ where $\frac{\partial}{\partial{t}}\phi(t,x)=f(\phi(t,x))$. We know by the chain rule that
\begin{align*}
\frac{\partial}{\partial{t}}\xi(\phi(t,x))&=\frac{\partial\xi}{\partial{x}}(\phi(t,x))\cdot \frac{\partial}{\partial{t}}\phi(t,x)
\\
&=\frac{\partial\xi}{\partial{x}}(\phi(t,x))\cdot f(\phi(t,x))
\\
&=\frac{\partial\xi}{\partial{x}}(x(t))\cdot f(x(t))
\end{align*}
where $\dfrac{\partial\xi(x)}{\partial{x}}\cdot{f(x)}$ is the directional derivative of the function $\xi$ in the direction of $f$. This is defined as
$$\dfrac{\partial\xi(x)}{\partial{x}}\cdot{f(x)}=\nabla{\xi(x)}\boldsymbol{\cdot}f(x)=\sum_{n=1}^{N}\frac{\partial{\xi}}{\partial{x_i}}f_i(x)$$
Therefore,
\begin{align*}
\frac{\partial}{\partial{t}}\xi(\phi(t,x))&=\frac{\partial\xi}{\partial{x}}(x(t))\cdot f(x(t))\\
&=\sum_{i=1}^n \frac{\partial\xi}{\partial{x_i}}(x(t)) f_i(x(t))
\\&=\sum_{i=1}^n f_i(\phi(t,x))\frac{\partial\xi}{\partial{x_i}}(\phi(t,x))
=L_{\phi(t,x)}[\xi]
\end{align*}
So along a solution we get $\frac{\partial}{\partial{t}}=L_{\phi(t,x)}$. Next, observe that we can write $x(a)$ in series notation as
$$x(a)=\sum_{n=0}^\infty\frac{x^{(n)}}{n!}a^n$$
We also know that
$$\exp(a)=\sum_{n=0}^\infty\frac{a^n}{n!}$$
Applying $\exp(a)$ to the differential operator $a\frac{\partial}{\partial{t}}$ produces
$$\exp\Big(a\frac{\partial}{\partial{t}}\Big)=\sum_{n=0}^\infty\frac{a^n}{n!}\Big(\frac{\partial}{\partial{t}}\Big)^n$$
Hence if we evaluate this at $x(t)$ we have
$$\exp\Big(a\frac{\partial}{\partial{t}}\Big)x(t)=\sum_{n=0}^\infty\frac{a^n}{n!}\Big(\frac{\partial{x(t)}}{\partial{t}}\Big)^n =\sum_{n=0}^\infty\frac{a^n}{n!}x^{n}(t)=\sum_{n=0}^\infty\frac{x^{n}(t)}{n!}a^n$$
Where it is necessary for $t=0$ in order to compute the Maclaurin expansion. Therefore,
$$\exp\Big(a\frac{\partial}{\partial{t}}\Big)x(t)\Big|_{t=0}=\sum_{n=0}^\infty\frac{x^{n}(t)}{n!}a^n=x(a)$$
Noting that $x(t)=\phi(t,x)$ and changing the constants $a\rightarrow{t}$, $t\rightarrow{s}$ so that $x(a)=x(t)$ and $\exp\Big(a\frac{\partial}{\partial{t}}\Big)x(t) = \exp\Big(t\frac{\partial}{\partial{s}}\Big)x(s)$ allows us to write
$$
\phi(t,x)=\exp\left(t\frac{\partial}{\partial{s}}\right)\phi(s,x)\Big|_{s=0}
=\exp\left(tL_{\phi(s,x)}\right)[\phi(s,x)]\Big|_{s=0}
=\exp\left(tL_{x}\right)[x]
$$
If we then replace the exponential with the exponential series, we have
$$
\phi(t,x)=\exp\left(tL_{x}\right)[x]=\Big(\sum_{n = 0}^{\infty} \frac{\left(tL_{x}\right)^n}{n!}\Big)[x]=\sum_{n=0}^{\infty}\frac{t^n}{n!}L^n[x]$$
