# Darboux's Theorem Alternate Proof

I've been given the task of proving Darboux's theorem through non-standard means.

Definitions Let $(M,\phi)$ be a symplectic manifold.

$\mathcal{F}_{\text{SP}(V)}(M)$ is the bundle of frames $(\eta^1,\ldots ,\eta^{2m})$ such that $\phi=\eta^1\wedge\eta^{m+1}+\cdots +\eta^m\wedge\eta^{2m}$.

$\eta$ is the column vector associated to a particular $(\eta^1,\ldots ,\eta^{2m})\in\mathcal{F}_{\text{SP}(V)}$, i.e. $\eta^i$ belongs to the $i^{\text{th}}$ row of $\eta$.

$\theta$ is a $\mathfrak{sp}(V)$-valued 1-form such that $d\eta=-\theta\wedge\eta$.

Problem

Let $(M,\phi)$ be a symplectic manifold. Compute $d\theta+\theta\wedge\theta$. Use this to prove Darboux's theorem.

I have an idea as how I might go about proving Darboux's theorem if I could compute $d\theta+\theta\wedge\theta$: I imagine (since $d\theta+\theta\wedge\theta$ is analogous to the Riemann curvature tensor) that it would be similar to showing that a Riemannian manifold is flat iff the Riemann curvature tensor is identically zero. But my progess halts at computing $d\theta+\theta\wedge\theta$. I cannot figure out anything besides breaking down $d\theta+\theta\wedge\theta$ into its component equations: If $\theta=(\theta^i_j)$ then solving $d\theta+\theta\wedge\theta$ is equivalent to solving $d\theta^i_j+\theta^i_k\wedge\theta^k_j$, but this doesn't get me anywhere. I've arrived a couple of equations using the fact that $d\phi=0$, but none of these have truly gotten me anywhere. Any help is greatly appreciated. Thanks in advance.

Edit: I forgot to mention it before, but this is a homework problem so just hints please. Thanks again.