I'm having trouble verifying an identity in symplectic geometry. Consider the cotangent bundle $T^*M$ over a smooth manifold $M$ where we can express coordinates locally as $(x,\xi)$. Let $\omega$ be the canonical symplectic form on this space where we can express $\omega = -d\alpha$ for the canonical 1-form, and $$ \omega \;\; =\;\; \sum_i dx^i\wedge d\xi^i \hspace{2pc}\text{and} \hspace{2pc} \alpha = \sum_i \xi_i dx^i = \left (d\pi_p\right )^*\xi $$
where $p =(x,\xi)$ and $d\pi_p:T_p\left (T^*M\right ) \to T_xM$
For any $h\in C^\infty(M)$ let $\tau_h:T^*M \to T^*M$ be the map given in local coordinates by $$ \tau_h(x,\xi) \;\; =\;\; (x, \xi + dh_x). $$ The goal is to prove the identity $$\tau_h^*\alpha \;\; =\;\; \alpha + \pi^*dh$$ where $\pi:T^*M \to M$ is the canonical projection. Once this is done, it is quite simple to show that $\tau_h^*\omega = \omega$, proving it to be a symplectomorphism.
I'm having a great deal of difficulty proving this claim. I've tried expressing the claim in terms of tangent vectors in $T_p(T^*M)$ but this got rather messy. I suspect there is a slick way to do this that I'm just missing.
The simplest approach is to prove the identity in Darboux coordinates $(x, \xi)$. Write $(x', \xi') = \tau_h(x, \xi) = (x, \xi + dh_x)$, hence $(x')^i = x^i$ and $\xi'_i = \xi_i + \frac{\partial h}{\partial x^i}(x)$. It follows that $$ \tau_h^*\alpha = \tau_h^*\left( \sum_i \xi_i dx^i \right) = \sum_i \xi'_i d(x')^i = \sum_i\left(\xi_i + \frac{\partial h}{\partial x^i}(x)\right) dx^i = \alpha + dh_x . $$