= Solution
A smooth <differential form> of degree $p$ is a smooth section of $\Lambda^pT^*M$, so
$$
\Omega^p(M)=\Gamma(\Lambda^pT^*M).
$$
It is equivalently a smooth alternating covariant $p$-tensor field. In a <coordinate chart> $x^1,\ldots,x^n$ it has a unique expression
$$
\omega=\sum_{i_1<\cdots<i_p}a_{i_1\cdots i_p}\,dx^{i_1}\wedge\cdots\wedge dx^{i_p}
$$
with smooth coefficients. At $p=0$ these are the <smooth functions>, and for $p>n$ the space is zero.
Define the <exterior derivative> in a chart by
$$
d\omega=\sum_{I,j}\frac{\partial a_I}{\partial x^j}\,dx^j\wedge dx^I.
$$
To check that it is well-defined, let $y^1,\ldots,y^n$ be other smooth coordinates. The expression $dy^a=\sum_i(\partial_i y^a)dx^i$ satisfies
$$
d_x(dy^a)=\sum_{i,j}(\partial_j\partial_i y^a)\,dx^j\wedge dx^i=0
$$
because the coefficients are symmetric and the wedge products antisymmetric. The coordinate formula also satisfies the <graded Leibniz rule>, and for a function $b$ the <chain rule> gives $d_xb=\sum_a(\partial b/\partial y^a)dy^a=d_yb$. Applying these two facts to $\omega=\sum_I b_I\,dy^I$ yields $d_x\omega=\sum_I d_yb_I\wedge dy^I=d_y\omega$. Thus the local operators agree on overlaps and define a global <linear map> $d:\Omega^p(M)\to\Omega^{p+1}(M)$.
A second application in any chart gives
$$
d^2\omega=\sum_{I,j,k}(\partial_k\partial_j a_I)\,dx^k\wedge dx^j\wedge dx^I=0,
$$
again by symmetry of mixed derivatives and antisymmetry of the <wedge product of differential forms>. Consequently every <exact differential form> is closed. Put $\Omega^{-1}(M)=0$ and define the <de Rham cohomology> by
$$
\boxed{H^p_{\mathrm{dR}}(M)=\frac{\ker(d:\Omega^p\to\Omega^{p+1})}{\operatorname{im}(d:\Omega^{p-1}\to\Omega^p)}.}
$$
The identity $d^2=0$ is exactly what puts the denominator inside the numerator, making this quotient meaningful. Since $\Omega^p(M)=0$ for $p>\dim M$, \b[$H^p_{\mathrm{dR}}(M)=0$ in those degrees].
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