Question about n-forms and $d^2 = 0$ In my late question the answer was great, but I didn't understand what is the 0-form, 1-form and 2-form. Are this can be genralized for any natural $n$?

they are both special cases of the fact that $d^2=0$

What this formula mean?
 A: Linear $n$-form is a function $L$ which maps $n$ vectors from $\mathbb{R}^m$ to a scalar $L(v_1,v_2,\dots,v_n) = a \in \mathbb R$ and has the linearity property: $$L(\alpha v_1 + \beta u_1, v_2, \dots, v_n) = \alpha L(v_1, v_2, \dots, v_n) + \beta L(u_1, v_2, \dots, v_n),$$
and so on for each $n$.
An alternating $n$-form is a linear form $L(v_1,\dots,v_n)$ such that if two vector arguments are equal, then $L=0$. For example, an alternating $2$-form is a linear form $L(v_1,v_2)$ such that $L(v,v)=0$ for every $v \in \mathbb{R}^3$.
If $v_1=(x_1,y_1,z_1)$ and $v_2 = (x_2,y_2,z_2)$ we use following denotations for some basic $n$-forms: $dx(v_1) = x_1$ is a $1$-form; $dx \wedge dy(v_1,v_2) = x_1 y_2 - x_2 y_1$ is an alternating $2$-form. Similarly we have $1$-forms $dy,dz$ and $2$-forms $dy \wedge dz, dz \wedge dx$.
If $\alpha$ and $\beta$ are $1$-forms, we define their wedge product $$(\alpha \wedge \beta) (v_1,v_2) = \alpha(v_1) \beta(v_2) - \alpha(v_2) \beta(v_1),$$
which is an alternating $2$-form. It's easily seen that $\alpha \wedge \beta = - \beta \wedge \alpha$ and $\alpha \wedge \alpha = 0$. The wedge product can be generalized to arbitrary forms: if $\alpha$ is an $m$-form and $\beta$ is an $n$-form, then $\alpha \wedge \beta$ is an alternating $(m+n)$-form. The generalization is a bit technical, and for now it's sufficient for you to know that the wedge product $\wedge$ is associative and anti-commutative.
A differential $n$-form (the type of form you are interested in) is made when you associate an alternating $n$-form to every point $(x,y,z) \in \mathbb{R}^3$ in a smooth way. For example, if $f,g$ and $h$ are smooth functions, then
$$ \omega_1 = f(x,y,z) dx + g(x,y,z) dy + h(x,y,z) dz $$
is a differential $1$-form,
$$ \omega_2 = f(x,y,z) dx \wedge dy + g(x,y,z) dy \wedge dz + h(x,y,z) dz \wedge dx $$
is a differential $2$-form and
$$ \omega_3 = f(x,y,z) dx \wedge dy \wedge dz$$
is a differential $3$-form. A differential $0$-form is defined to be a scalar field on $\mathbb{R}^3$. When working on $\mathbb{R}^3$ we are only interested in $0,1,2$ and $3$ forms, since for $n \geq 4$ every $n$-form is equal to $0$. 
The exterior derivative of a form takes an $n$-form to an $(n+1)$-form. If $f$ is a $0$-form (that is a scalar field) then the exterior derivative is the same thing as the total differential:
$$ df = f_x dx + f_y dy + f_z dz $$
where $f_{x_i}$ denotes the partial derivative. If $\omega = f dx$ is a $1$-form, then
$$d \omega = d f \wedge dx = (f_x d x + f_y d y + f_z dz) \wedge dx = - f_y dx \wedge d y + f_z dz \wedge dx$$
(bear in mind that $d x \wedge dx = 0$). And for $\omega = f dx \wedge dy$ a two form:
$$d \omega = d f \wedge dx \wedge dy = f_z dz \wedge dx \wedge d y = f_z dx \wedge dy \wedge dz.$$
Now you can check for yourself that for every form $\omega$, $d^2 \omega = d(d \omega) = 0$.
Differential forms can be generalized to every $n$-dimensional differentiable manifold. We can develop integration theory for differential forms which leads us to a generalized Stokes' theorem and de Rham's theorem.
In $\mathbb{R}^3$ is a natural correspondence between vector fields and $1$-forms (or $2$-forms). To every vector field $V(x,y,z) = (V_1, V_2, V_3)$ we can associate a $1$-form
$$\omega_{1,V} = V_1 dx + V_2 d_y + V_3 dz$$
and a $2$-form
$$\omega_{2,V} = V_1 dy \wedge dz + V_2 dz \wedge dx + V_3 dx \wedge dy.$$
If $U = \mathrm{curl}\, V$, then $\omega_{2,U} = d \omega_{1,V}$. If $f = \mathrm{div}\, V$ then $d \omega_{2,V} = f dx \wedge dy \wedge dz$.
Hope this helps.
A: $d^2 = 0$ is a restatement of the equality of mixed partial derivatives.
Differential forms may be more easily understood in terms of a geometric approach.  In an N-dimensional real vector space, one typically only deals with vectors and scalars.  Scalars form their own 0-dimensional "vector" space, as all elements are just multiples of each other.  Vectors describe various directions, or lines through the origin that are oriented--pointing one way or another.
But there is more to the world than just scalars and vectors.  How would you describe, for example, a plane through the origin?  There is an equation for a plane, yes, but in terms of vectors?  In 3D, each plane has a unique normal vector, but for $N$ dimensions, that is no longer the case.
The solution to this problem is the wedge product, which is anticommutative (like the cross product), but the wedge product of two vectors is not another vector.  Rather, it is called a bivector.
Let me be concrete and talk about a 3D vector space.  Let there be three basis vectors, $e_x, e_y, e_z$, corresponding to the $xyz$ axes, respectively.
In this space, one has scalars, which form their own 0-dimensional vector space.
One also has vectors, which are linear combinations of the three basis vectors. A direction could be given as $x e_x + y e_y + z e_z$, for instance.
One has bivectors, which are oriented planes in the space.  The basis planes are $e_x \wedge e_y, e_x \wedge e_z, e_y \wedge e_z$--3 in total.  Note that $e_i \wedge e_i = 0$ for any of the three dimensions.  It's interesting that there are only 3 basis planes in 3D.  This illustrates how each plane has a unique normal vector, but again, in other dimensions such a relationship doesn't hold.  This relationship between bivectors and vectors in 3d is called duality.
Finally, there is a single trivector, also called the pseudoscalar because it forms a one-dimensional vector space like the scalars but is negated on reflections.  The basis pseudoscalar is $e_x \wedge e_y \wedge e_z$.  The pseudosclar corresponds to an oriented volume.
In 3D, this can seem like a lot of extra bookkeeping--instead of just scalar and vector fields, we have bivector and trivector (pseudoscalar) fields, but the ease with which this approach generalizes to any number of dimensions makes it useful, and the additional complexity isn't too bad.
Now that we've established the framework of multivectors, we can talk about $\nabla$.  It takes on a similar role to what you're used to in ordinary vector calculus.  $\nabla = e^x \partial_x + e^y \partial_y + e^z \partial_z$.  We can form two derivative operators on a vector field $A$ with this symbol--$\nabla \cdot A$ uses the familiar dot product and produces a scalar field.  $\nabla \wedge A$ uses the new wedge product and produces a bivector field, but it is very similar to the curl.  Both of these operators have a unique property:
$$\nabla \cdot \nabla \cdot A = \nabla \wedge \nabla \wedge A = 0$$
Differential forms parlance could call $\nabla \cdot A \equiv \delta A$ and $\nabla \wedge A\equiv dA$, but it's all really saying the same thing.
What does this mean?  Well, $\nabla \cdot \nabla \cdot A = 0$ is trivial.  Once you have $\nabla \cdot A$, a scalar field, you can't dot product it with anything anymore.  $\nabla \wedge \nabla \wedge A$, however, is interesting.  Let's expand that in terms of components.  Let $A = A_x e^x + A_y e^y + A_z e^z$, and we get
$$\nabla \wedge \nabla \wedge A = (e^x \wedge e^y \wedge e^z) \left[(\partial_x \partial_y - \partial_y \partial_x) A_z + (\partial_z \partial_x -\partial_x \partial_z) A_y + (\partial_y \partial_z - \partial_z \partial_y) A_x \right]$$
There are a lot of mixed partial derivatives here, and what do we know about mixed partials?  They're equal!  So all these terms cancel each other out, and you get 0.  $d^2 = 0$ is a restatement of the equality of mixed partial derivatives.
