Cotangent Fields: Exactness vs. Conservation Given a smooth manifold. Then a cotangent field is exact iff conservative: $$\alpha\in\mathcal{X}^*(M):\quad\alpha=\mathrm{d}h\iff\oint\alpha=0$$ How to prove this properly?

_This proof is taken from Lee's Smooth Manifolds._

Consider up to a constant: $$h(x)=\int_z^x\alpha$$

Regard a smooth coordinate path: $$\gamma:[0,\varepsilon)\to M:\quad\hat{\gamma}(t):=(0,\ldots,0,t,0,\ldots,0)\quad(\gamma(0)=z)$$ So one obtains as partial derivatives: $$\frac{\partial h}{\partial x^i}(z)=\gamma'(0)h=\left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}(h\circ\gamma)=\left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}\int_z^{\gamma(t)}\alpha\\\=\left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}\int_0^t\alpha(\gamma'(s))\mathrm{d}s=\alpha(\gamma'(0))=\alpha^i(z)$$ _(This especially shows smoothness.)_