Poincare lemma states that every smooth closed form in contractible subset is exact. The assumption actually represents a set of pdes that dd=0. Could Frobenius theorem be used to say that the solution of closed form is exact form?
See Bott en Tu's wonderful book "differential forms in algebraic topology". They explain clearly that the "right" formulation is : every k-form ($\omega$ on a domain $D$ which is closed (i.e. $d \omega = 0$) is exact (i.e. $\omega = d \theta$, for some (k-1) if and only if the k-th cohomology with real coefficients $H^k(D, R) = 0$. This is known as de Rham's theorem and relates a topological condition to the solutions of a pde. The topological condition is certainly satisfied if the domain D is contractible.
If you are into numerical mathematics you may also want to have a look at Arnold, Falk, Winther, acta Numerica (2006) .
That said, the proof is really easy and explicit if D is smoothly contractible. Lets assume $ k > 0$ since for k = 0, de Rham's theorem just says that a smooth function $f$ with $df = 0$ is locally constant (and the proof below works mutatis mutandis).
Let $I = [0,1]$ and $f: D\times I \to D$ be a smooth contraction i.e. $f(x, 0) = x$ en $f(x, 1) = P_0$ for some point $P_0$ in $D$. Denote the exterior derivative on a smooth manifold like $D$ by $d_D$. Then if $\omega$ is a closed smooth $ k$ form on $D$ (i.e $d\omega = d_D \omega = 0$) we have
where in the last line we used that $f*\omega|_{t = 1}$ is the restriction of a $k$ form to the point $P_0$ and so vanishes for $k > 0$ (or is a constant for $k = 0$).
I don't think it follows from the Frobenius theorem, because it is a statement about how solutions of a pde are determined by the global topology of you domain. You are right however that the Frobenius theorem is closely related to the fact that a closed 1 form is exact in a neighborhood of a point in a manifold