Writing Exterior Derivative in terms of Lie Bracket let $\alpha\in\Omega^1(U, \mathbb{R})$ be a 1-form on $U$ and $X, Y : U → \mathbb{R}^n$ vector fields.
Verify that the exterior derivative can be calculated via
$$d\alpha(X, Y ) = X \cdot\alpha(Y ) − Y \cdot \alpha(X) − \alpha([X, Y ])$$
where the smooth function $\alpha(X): U → \mathbb{R}$  is given by $$\alpha(X)(p) := \alpha_p(X(p))$$
for $p\in U$.
I haven't seen a ton on the exterior derivative and the bit that I have has involved sort of how it operates in the abstract on the cochain complex. In other words showing how it moves a 1-form to a 2 form and so on..
My instinct would be to rewrite the equation for the exterior derivative using the definition of the function $\alpha$(X) and the definition of the lie bracket and see where that leads me. Or alternatively, to rewrite the equation in order to equate the lie bracket(multiplied by $\alpha$ (the one form)) to the remaining terms and hope that they work out nicely. Any help?
 A: Since $d\alpha$ is a differential form, and in particular a tensor, the value of $d\alpha(X,Y)$ at the point $p$ only depends on the values of $X$ and $Y$ at $p$. Hence, you may assume that $X$ and $Y$ are constant vector fields. This makes the whole computation much simpler.
A: Old question but maybe the proof helps someone either way.
So let $\omega \in \Omega^k(U)$ where $U$ is an open neighborhood of $\mathbb{R}^n$. Let further $X,Y \in \Gamma(U)$ denote smooth vector fields over $U$. Without loss of generality we can locally express $\omega$ by $\omega = f dg$ using smooth functions $f,g \in C^\infty (U)$. This allows us to calculate
\begin{align*}
d \omega(X,Y) &= d(f dg) (X,Y) = df \wedge dg (X,Y)\\\\
              &= df(X) dg(Y) - df(Y) dg(X)\\\\
              &= X(f) Y(g) - Y(f) X(g)
\end{align*}
Where in the last equation we used that for smooth functions $df(X) = X(f)$. As we want to incorporate the Lie-bracket in the final result, we go on to add a rather blatant zero:
\begin{align*}
d \omega(X,Y) &= X(f) Y(g) - Y(f) X(g) - f[ X(Y(g)) - Y(X(g))] + f[X(Y(g))] - f[Y(X(g))]\\\\
              &= X(f)Y(g) + f X(Y(g)) - Y(f)X(g) - f(Y(X(g)) - \omega([X,Y])\\\\
              &= X(\omega(Y)) - Y(\omega(X)) - \omega([X,Y])
\end{align*}
Due to $f[ X(Y(g)) - Y(X(g))] + f[X(Y(g))] = \omega([X,Y])$ the first equality holds. Regrouping then yields the desired result.
