Proof: For ${\displaystyle k\in \mathbb {N} \cup \{0\}}$, it is clear that ${\displaystyle d}$ maps ${\displaystyle \Omega ^{k}(M)}$ to ${\displaystyle \Omega ^{k+1}(M)}$. We claim that also ${\displaystyle d\circ d=0}$. By linearity, we reduce to the case of a basis element, so suppose that ${\displaystyle \omega =fdx_{i_{1}}\wedge \cdots \wedge dx_{i_{k}}\in \Omega ^{k}(M)}$ with ${\displaystyle i_{1}<\cdots <i_{k}}$ and ${\displaystyle f\in C^{\infty }(M)}$. Then

${\displaystyle {\begin{aligned}(d\circ d)(\omega )&=d\left(\sum _{j=1}^{n}{\frac {\partial f}{\partial x_{j}}}dx_{j}\wedge dx_{i_{1}}\wedge \cdots \wedge dx_{i_{k}}\right)\\&=\sum _{k,j=1}^{n}{\frac {\partial ^{2}f}{\partial x_{j}\partial x_{k}}}dx_{k}\wedge dx_{j}\wedge dx_{i_{1}}\wedge \cdots \wedge dx_{i_{k}}.\end{aligned}}}$

By Clairaut's theorem and the anti-commutativity of ${\displaystyle \wedge }$, all terms cancel except the ones where ${\displaystyle k=j}$, and there ${\displaystyle dx_{k}\wedge dx_{j}=0}$. ${\displaystyle \Box }$