[proofplan] Apply Taylor's theorem to $f(\tilde{x}(\varepsilon))$ where $\tilde{x}(\varepsilon) = \psi^\varepsilonx$ is the flow. The chain rule shows that each successive $\varepsilon$-derivative at $\varepsilon = 0$ equals applying the differential operator $V$ one more time, giving the $n$-th coefficient as $V^n f / n!$. [/proofplan] [step:Expand $f(\psi^\varepsilonx)$ by Taylor's theorem in $\varepsilon$] By Taylor's theorem applied to the smooth function $\varepsilon \mapsto f(\tilde{x}(\varepsilon))$: \begin{align*} f(\psi^\varepsilonx) = \sum_{n=0}^\infty \frac{\varepsilon^n}{n!}\left.\frac{d^n}{d\varepsilon^n}f(\tilde{x}(\varepsilon))\right|_{\varepsilon = 0}. \end{align*} It remains to show that the $n$-th $\varepsilon$-derivative at $\varepsilon = 0$ equals $(V^n f)(x)$. [/step] [step:Show $\frac{d}{d\varepsilon}f(\tilde{x}(\varepsilon)) = (Vf)(\tilde{x}(\varepsilon))$ by the chain rule] Since $\tilde{x}(\varepsilon)$ solves $d\tilde{x}/d\varepsilon = V(\tilde{x})$, the chain rule gives \begin{align*} \frac{d}{d\varepsilon}f(\tilde{x}(\varepsilon)) = \sum_{i=1}^n \frac{d\tilde{x}_i}{d\varepsilon}\frac{\partial f}{\partial x_i}\bigg|_{\tilde{x}(\varepsilon)} = \sum_{i=1}^n V_i(\tilde{x})\frac{\partial f}{\partial x_i}\bigg|_{\tilde{x}(\varepsilon)} = (Vf)(\tilde{x}(\varepsilon)). \end{align*} [/step] [step:Iterate to show the $n$-th derivative equals $V^n f$ at $\varepsilon = 0$] Applying the result of the previous step to $Vf$ in place of $f$: \begin{align*} \frac{d^2}{d\varepsilon^2}f(\tilde{x}(\varepsilon)) = \frac{d}{d\varepsilon}(Vf)(\tilde{x}(\varepsilon)) = (V(Vf))(\tilde{x}(\varepsilon)) = (V^2 f)(\tilde{x}(\varepsilon)). \end{align*} By induction, $\frac{d^n}{d\varepsilon^n}f(\tilde{x}(\varepsilon)) = (V^n f)(\tilde{x}(\varepsilon))$ for all $n \geq 0$. Evaluating at $\varepsilon = 0$ (where $\tilde{x}(0) = x$): \begin{align*} \left.\frac{d^n}{d\varepsilon^n}f(\tilde{x}(\varepsilon))\right|_{\varepsilon = 0} = (V^n f)(x). \end{align*} Substituting into the Taylor expansion: \begin{align*} f(\psi^\varepsilonx) = \sum_{n=0}^\infty \frac{\varepsilon^n}{n!}(V^n f)(x) = (\exp(\varepsilon V)f)(x). \end{align*} [/step]