= Solution
The space of smooth <differential forms> of degree $p$ is
$$
\boxed{\Omega^p(M)=\Gamma(\Lambda^pT^*M).}
$$
Thus a form assigns smoothly to each point an alternating $p$-linear function on its <tangent space>. Here $\Omega^0(M)=C^\infty(M)$, and forms of degree greater than $\dim M$ vanish. The <wedge product of differential forms> gives the graded <exterior algebra> of forms.
The <exterior derivative> is an $\mathbb R$-linear map of degree one satisfying the following axioms: $df(X)=X(f)$ for functions; for a $p$-form $\alpha$,
$$
d(\alpha\wedge\beta)=d\alpha\wedge\beta+(-1)^p\alpha\wedge d\beta;
$$
and $d^2=0$. The sign is the <graded Leibniz rule>.
To construct it, write in a <coordinate chart>
$$
\alpha=\sum_{i_1<\cdots<i_p}a_{i_1\ldots i_p}\,
dx^{i_1}\wedge\cdots\wedge dx^{i_p},
\qquad
d\alpha=\sum_{i_1<\cdots<i_p}da_{i_1\ldots i_p}\wedge
dx^{i_1}\wedge\cdots\wedge dx^{i_p}.
$$
This formula is linear, agrees with differentiation on functions and obeys the <graded Leibniz rule>. Its square vanishes because symmetric second partial derivatives are contracted against antisymmetric wedge products.
The construction is independent of the chart. For coordinates $y^j=y^j(x)$, the <chain rule> gives $dy^j=\sum_i(\partial_i y^j)dx^i$, and the coordinate formula gives $d(dy^j)=0$ by equality of mixed partial derivatives. Consequently, expressing a form as $\sum_I b_I\,dy^I$ and applying the $x$-coordinate formula gives exactly $\sum_I db_I\wedge dy^I$, the $y$-coordinate formula. The local constructions agree on overlaps and therefore define a global <exterior derivative>.
For uniqueness, first derive locality from the axioms. If $\alpha$ vanishes near $p$, choose a <smooth bump function> $\chi$ supported in that neighborhood and equal to one near $p$. Then $\chi\alpha=0$, and
$$
0=d(\chi\alpha)=d\chi\wedge\alpha+\chi\,d\alpha
$$
implies $(d\alpha)_p=0$. Thus the value depends only on the local germ of $\alpha$. Local coordinate functions can be extended using bump functions, so the same axioms apply to their germs. Since $d(dx^i)=d^2x^i=0$, the <graded Leibniz rule> forces precisely the coordinate formula above. \b[The exterior derivative exists and is unique], as expressed by <uniqueness of the exterior derivative from its axioms>.
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