TITLE: Coordinate Free Proof of $ d^2 = 0$
QUESTION [6 upvotes]: We all know that when the exterior derivative is applied twice to a $k$ form, it always yields the null $k+2$ form (i.e. $d^2 = 0$). However, the only proofs I've seen of this does it in some chart and using coordinate notation computes this. I was wondering if anyone has seen or knows of a proof of $d^2 = 0$ that doesn't use coordinates and instead proves this in a coordinate free way.

REPLY [2 votes]: I found a self-contained answer that really requires no real computation. I'll reproduce a sketch of it below but it can be found in Natural Operations in Differential Geometry on pg. 66. I think the key is recognizing what generates the algebra $ \Omega(M) $ locally at least. 
Since $d$ is a graded derivation of degree one, $ d^2 = \frac12[d,d] $ is a graded derivation of degree two and this is local. And since $\Omega(M) $ is locally generated by $ C^\infty(M,\mathbb{R} )$ and $ \{ df: f\in C^\infty(M,\mathbb{R})\} $ you can just show $ d^2f=0$. By simple computation
$$ d^2f(X,Y)= Xdf(Y)-Ydf(X)-df([X,Y])=XYf-YXf-[X,Y]f = 0 $$
Therefore $d^2 = 0$ and $ d^l=0$ for all $ l \geq 2$ for that matter.