[proofplan] The fundamental vector field $\xi_M$ is generated by the one-parameter subgroup $\exp(t\xi)$ acting on $M$. Since the action is symplectic, every time-$t$ action map preserves the symplectic form $\theta$, and differentiating this pullback identity gives $\mathcal L_{\xi_M}\theta=0$. Finally, Cartan's formula and the closedness of the symplectic form identify this vanishing condition with the closedness of $\iota_{\xi_M}\theta$. [/proofplan]

[step:Identify the flow of the fundamental vector field with the one-parameter group action]Fix $\xi\in\mathfrak g$. Define a smooth family of diffeomorphisms \begin{align*} \Phi_t:M\to M \end{align*} by \begin{align*} \Phi_t(x)=\exp(t\xi)\cdot x. \end{align*} By the definition of the fundamental vector field associated to a left action, \begin{align*} \frac{d}{dt}\bigg|_{t=0}\Phi_t(x)=\xi_M(x) \end{align*} for every $x\in M$. More generally, using the one-parameter subgroup identity $\exp((t+s)\xi)=\exp(t\xi)\exp(s\xi)$ and the left action law, the curve $t\mapsto \Phi_t(x)$ has velocity $\xi_M(\Phi_t(x))$. Hence $(\Phi_t)$ is the flow of $\xi_M$.[/step]

[guided]We first convert the infinitesimal object $\xi_M$ into a flow, because the Lie derivative of a form is computed by differentiating pullbacks along a flow. For the fixed element $\xi\in\mathfrak g$, define \begin{align*} \Phi_t:M\to M \end{align*} by \begin{align*} \Phi_t(x)=\exp(t\xi)\cdot x. \end{align*} This is a smooth family of diffeomorphisms because each $\exp(t\xi)\in G$ acts on $M$ by a diffeomorphism. By the chosen convention for the fundamental vector field, \begin{align*} \xi_M(x)=\frac{d}{dt}\bigg|_{t=0}\exp(t\xi)\cdot x. \end{align*} Thus the initial velocity of the curve $t\mapsto \Phi_t(x)$ is exactly $\xi_M(x)$. To check that $\Phi_t$ is the full flow, not only the correct initial velocity at $t=0$, fix $s\in\mathbb R$. The velocity at time $s$ is computed by shifting the parameter: \begin{align*} \frac{d}{dt}\bigg|_{t=0}\Phi_{s+t}(x)=\frac{d}{dt}\bigg|_{t=0}\exp((s+t)\xi)\cdot x. \end{align*} Since $\xi$ generates a one-parameter subgroup, $\exp((s+t)\xi)=\exp(t\xi)\exp(s\xi)$. Therefore \begin{align*} \frac{d}{dt}\bigg|_{t=0}\Phi_{s+t}(x)=\frac{d}{dt}\bigg|_{t=0}\exp(t\xi)\cdot(\exp(s\xi)\cdot x). \end{align*} The right-hand side is, by definition of $\xi_M$, equal to \begin{align*} \xi_M(\Phi_s(x)). \end{align*} Hence the curve $s\mapsto\Phi_s(x)$ solves the differential equation for the vector field $\xi_M$, so $(\Phi_t)$ is the flow of $\xi_M$.[/guided]

[step:Differentiate the symplectic invariance identity] Because the action of $G$ on $(M,\theta)$ is symplectic, each diffeomorphism $\Phi_t$ preserves $\theta$. Thus \begin{align*} \Phi_t^*\theta=\theta \end{align*} for every $t$ for which the flow is defined. By the defining flow formula for the Lie derivative of a differential form, \begin{align*} \mathcal L_{\xi_M}\theta=\frac{d}{dt}\bigg|_{t=0}\Phi_t^*\theta. \end{align*} Differentiating the constant family $\Phi_t^*\theta=\theta$ gives \begin{align*} \mathcal L_{\xi_M}\theta=0. \end{align*} [/step]

[step:Use Cartan's formula to translate vanishing Lie derivative into closedness] Cartan's formula for the Lie derivative of a differential form gives \begin{align*} \mathcal L_{\xi_M}\theta=d(\iota_{\xi_M}\theta)+\iota_{\xi_M}(d\theta). \end{align*} Since $\theta$ is a symplectic form, it is closed, so \begin{align*} d\theta=0. \end{align*} Therefore \begin{align*} \mathcal L_{\xi_M}\theta=d(\iota_{\xi_M}\theta). \end{align*} The equality $\mathcal L_{\xi_M}\theta=0$ is consequently equivalent to \begin{align*} d(\iota_{\xi_M}\theta)=0. \end{align*} This is precisely the statement that the one-form $\iota_{\xi_M}\theta$ is closed. [/step]