[proofplan]
We verify each identification by direct coordinate computation. The gradient case follows immediately from the coordinate formula for $df$. For the curl and divergence cases, we expand $d\alpha_F$ and $d\beta_F$ using the antisymmetry of the wedge product and collect coefficients in the standard $2$-form and $3$-form bases. The vector calculus identities $\nabla \times \nabla f = 0$ and $\nabla \cdot (\nabla \times F) = 0$ then follow by applying $d$ twice and invoking the nilpotency $d^2 = 0$ of the [Exterior Derivative](/theorems/1525).
[/proofplan]
[step:Identify $df$ with the gradient $\nabla f$]
By the [coordinate formula for the exterior derivative](/theorems/3564) of a smooth $0$-form (see the [Exterior Derivative](/theorems/1525)),
\begin{align*}
df = \frac{\partial f}{\partial x_1}\, dx_1 + \frac{\partial f}{\partial x_2}\, dx_2 + \frac{\partial f}{\partial x_3}\, dx_3.
\end{align*}
The components of the gradient are $(\nabla f)_i = \partial_{x_i} f$ for $i = 1, 2, 3$, so
\begin{align*}
\alpha_{\nabla f} = \sum_{i=1}^{3} (\nabla f)_i\, dx_i = \sum_{i=1}^{3} \frac{\partial f}{\partial x_i}\, dx_i = df.
\end{align*}
[guided]
We start with the most basic case: the [exterior derivative](/theorems/1525) of a smooth function (a $0$-form). By definition, for $f \in C^\infty(U)$ the [exterior derivative](/theorems/1525) is the $1$-form
\begin{align*}
df = \sum_{i=1}^{3} \frac{\partial f}{\partial x_i}\, dx_i,
\end{align*}
which is exactly the coordinate formula from the [Exterior Derivative](/theorems/1525) specialised to $n = 3$.
Now recall how we built $\alpha_F$ from a vector field $F$: we took the components $F_i$ and made them the coefficients of $dx_i$. So $\alpha_F = \sum_i F_i\, dx_i$. Applying this construction to the vector field $\nabla f$, whose $i$-th component is $\partial_{x_i} f$, we obtain
\begin{align*}
\alpha_{\nabla f} = \sum_{i=1}^{3} (\nabla f)_i\, dx_i = \sum_{i=1}^{3} \frac{\partial f}{\partial x_i}\, dx_i.
\end{align*}
The right-hand side is exactly $df$, so $df = \alpha_{\nabla f}$. In words: under the metric identification of $1$-forms with vector fields, the [exterior derivative](/theorems/1525) on $0$-forms is the gradient.
[/guided]
[/step]
[step:Identify $d\alpha_F$ with the curl $\nabla \times F$]
Expanding $d\alpha_F$ component-by-component using $d(g\, dx_i) = dg \wedge dx_i$ and $dx_i \wedge dx_i = 0$:
\begin{align*}
d\alpha_F &= \sum_{i=1}^{3} dF_i \wedge dx_i = \sum_{i=1}^{3} \sum_{j=1}^{3} \frac{\partial F_i}{\partial x_j}\, dx_j \wedge dx_i \\
&= \sum_{j < i} \left( \frac{\partial F_i}{\partial x_j} - \frac{\partial F_j}{\partial x_i} \right) dx_j \wedge dx_i,
\end{align*}
where we used $dx_i \wedge dx_j = -dx_j \wedge dx_i$ to collect ordered pairs. Writing out the three terms:
\begin{align*}
d\alpha_F = \left( \frac{\partial F_2}{\partial x_1} - \frac{\partial F_1}{\partial x_2} \right) dx_1 \wedge dx_2 + \left( \frac{\partial F_3}{\partial x_2} - \frac{\partial F_2}{\partial x_3} \right) dx_2 \wedge dx_3 + \left( \frac{\partial F_3}{\partial x_1} - \frac{\partial F_1}{\partial x_3} \right) dx_1 \wedge dx_3.
\end{align*}
Using $dx_1 \wedge dx_3 = -dx_3 \wedge dx_1$ to align with the basis $(dx_2 \wedge dx_3, dx_3 \wedge dx_1, dx_1 \wedge dx_2)$:
\begin{align*}
d\alpha_F = \left( \frac{\partial F_3}{\partial x_2} - \frac{\partial F_2}{\partial x_3} \right) dx_2 \wedge dx_3 + \left( \frac{\partial F_1}{\partial x_3} - \frac{\partial F_3}{\partial x_1} \right) dx_3 \wedge dx_1 + \left( \frac{\partial F_2}{\partial x_1} - \frac{\partial F_1}{\partial x_2} \right) dx_1 \wedge dx_2.
\end{align*}
The components of the curl in $\mathbb{R}^3$ are
\begin{align*}
(\nabla \times F)_1 &= \partial_{x_2} F_3 - \partial_{x_3} F_2, \\
(\nabla \times F)_2 &= \partial_{x_3} F_1 - \partial_{x_1} F_3, \\
(\nabla \times F)_3 &= \partial_{x_1} F_2 - \partial_{x_2} F_1,
\end{align*}
so $d\alpha_F = \beta_{\nabla \times F}$.
[guided]
We now compute $d\alpha_F$ where $\alpha_F = \sum_i F_i\, dx_i$. The Leibniz rule for the [exterior derivative](/theorems/1525) gives $d(F_i\, dx_i) = dF_i \wedge dx_i$ (since $d(dx_i) = 0$ on a $1$-form basis element). Therefore
\begin{align*}
d\alpha_F = \sum_{i=1}^{3} dF_i \wedge dx_i = \sum_{i=1}^{3} \sum_{j=1}^{3} \frac{\partial F_i}{\partial x_j}\, dx_j \wedge dx_i.
\end{align*}
The diagonal terms $j = i$ vanish because $dx_i \wedge dx_i = 0$. For the off-diagonal terms, we collect each ordered pair $\{j, i\}$ with $j < i$ exactly once, using $dx_i \wedge dx_j = -dx_j \wedge dx_i$ to align signs:
\begin{align*}
d\alpha_F = \sum_{j < i} \left( \frac{\partial F_i}{\partial x_j}\, dx_j \wedge dx_i + \frac{\partial F_j}{\partial x_i}\, dx_i \wedge dx_j \right) = \sum_{j < i} \left( \frac{\partial F_i}{\partial x_j} - \frac{\partial F_j}{\partial x_i} \right) dx_j \wedge dx_i.
\end{align*}
In $\mathbb{R}^3$ the three pairs $(j, i)$ with $j < i$ are $(1,2), (1,3), (2,3)$. Listing the corresponding terms:
\begin{align*}
d\alpha_F = \left( \frac{\partial F_2}{\partial x_1} - \frac{\partial F_1}{\partial x_2} \right) dx_1 \wedge dx_2 + \left( \frac{\partial F_3}{\partial x_1} - \frac{\partial F_1}{\partial x_3} \right) dx_1 \wedge dx_3 + \left( \frac{\partial F_3}{\partial x_2} - \frac{\partial F_2}{\partial x_3} \right) dx_2 \wedge dx_3.
\end{align*}
Now we want to compare to the standard $2$-form basis used in the definition of $\beta_F$, namely $(dx_2 \wedge dx_3, dx_3 \wedge dx_1, dx_1 \wedge dx_2)$. This basis is the cyclically-ordered one, oriented compatibly with the standard volume form $dx_1 \wedge dx_2 \wedge dx_3$. We have $dx_1 \wedge dx_3 = -dx_3 \wedge dx_1$, so the middle term flips sign:
\begin{align*}
d\alpha_F = \left( \frac{\partial F_3}{\partial x_2} - \frac{\partial F_2}{\partial x_3} \right) dx_2 \wedge dx_3 + \left( \frac{\partial F_1}{\partial x_3} - \frac{\partial F_3}{\partial x_1} \right) dx_3 \wedge dx_1 + \left( \frac{\partial F_2}{\partial x_1} - \frac{\partial F_1}{\partial x_2} \right) dx_1 \wedge dx_2.
\end{align*}
Reading off the coefficients, they are exactly $(\nabla \times F)_1, (\nabla \times F)_2, (\nabla \times F)_3$ in turn. By the definition of $\beta_G$ applied to $G = \nabla \times F$, this is precisely $\beta_{\nabla \times F}$. The sign flip we just did is the entire content of why the curl uses a cyclic alternating sum: it is the algebraic shadow of the wedge antisymmetry $dx_1 \wedge dx_3 = -dx_3 \wedge dx_1$.
[/guided]
[/step]
[step:Identify $d\beta_F$ with the divergence $\nabla \cdot F$]
By the Leibniz rule and $d(dx_i \wedge dx_j) = 0$:
\begin{align*}
d\beta_F = dF_1 \wedge dx_2 \wedge dx_3 + dF_2 \wedge dx_3 \wedge dx_1 + dF_3 \wedge dx_1 \wedge dx_2.
\end{align*}
Expanding $dF_i = \sum_{j=1}^{3} \partial_{x_j} F_i\, dx_j$ and discarding terms in which a coordinate $1$-form repeats:
\begin{align*}
dF_1 \wedge dx_2 \wedge dx_3 &= \frac{\partial F_1}{\partial x_1}\, dx_1 \wedge dx_2 \wedge dx_3, \\
dF_2 \wedge dx_3 \wedge dx_1 &= \frac{\partial F_2}{\partial x_2}\, dx_2 \wedge dx_3 \wedge dx_1 = \frac{\partial F_2}{\partial x_2}\, dx_1 \wedge dx_2 \wedge dx_3, \\
dF_3 \wedge dx_1 \wedge dx_2 &= \frac{\partial F_3}{\partial x_3}\, dx_3 \wedge dx_1 \wedge dx_2 = \frac{\partial F_3}{\partial x_3}\, dx_1 \wedge dx_2 \wedge dx_3,
\end{align*}
where we cyclically permuted to the standard order $dx_1 \wedge dx_2 \wedge dx_3$ (each cyclic permutation of three elements is even). Summing:
\begin{align*}
d\beta_F = \left( \frac{\partial F_1}{\partial x_1} + \frac{\partial F_2}{\partial x_2} + \frac{\partial F_3}{\partial x_3} \right) dx_1 \wedge dx_2 \wedge dx_3 = (\nabla \cdot F)\, dx_1 \wedge dx_2 \wedge dx_3.
\end{align*}
[guided]
We now compute $d\beta_F$ for $\beta_F = F_1\, dx_2 \wedge dx_3 + F_2\, dx_3 \wedge dx_1 + F_3\, dx_1 \wedge dx_2$. Since $d$ annihilates the constant-coefficient $2$-forms $dx_i \wedge dx_j$, the Leibniz rule gives
\begin{align*}
d\beta_F = dF_1 \wedge dx_2 \wedge dx_3 + dF_2 \wedge dx_3 \wedge dx_1 + dF_3 \wedge dx_1 \wedge dx_2.
\end{align*}
For each $i$, write $dF_i = \sum_j \partial_{x_j} F_i\, dx_j$. The key observation is that in each term $dF_i \wedge (\text{the }2\text{-form not involving } dx_i)$, all summands $\partial_{x_j} F_i\, dx_j \wedge \cdots$ with $j \ne i$ contain a repeated factor and therefore vanish. Only the summand $j = i$ survives.
For the first term, the $2$-form is $dx_2 \wedge dx_3$, so only $j = 1$ survives:
\begin{align*}
dF_1 \wedge dx_2 \wedge dx_3 = \partial_{x_1} F_1\, dx_1 \wedge dx_2 \wedge dx_3.
\end{align*}
For the second term, the $2$-form is $dx_3 \wedge dx_1$, so only $j = 2$ survives:
\begin{align*}
dF_2 \wedge dx_3 \wedge dx_1 = \partial_{x_2} F_2\, dx_2 \wedge dx_3 \wedge dx_1.
\end{align*}
We want this in the standard order $dx_1 \wedge dx_2 \wedge dx_3$. The cyclic permutation $(2,3,1) \to (1,2,3)$ is an even permutation (it is the square of a transposition), so $dx_2 \wedge dx_3 \wedge dx_1 = dx_1 \wedge dx_2 \wedge dx_3$.
Similarly for the third term, with $(3,1,2) \to (1,2,3)$ also even:
\begin{align*}
dF_3 \wedge dx_1 \wedge dx_2 = \partial_{x_3} F_3\, dx_3 \wedge dx_1 \wedge dx_2 = \partial_{x_3} F_3\, dx_1 \wedge dx_2 \wedge dx_3.
\end{align*}
This is exactly why the $2$-form basis was chosen to be the cyclically ordered $(dx_2 \wedge dx_3,\, dx_3 \wedge dx_1,\, dx_1 \wedge dx_2)$ rather than the lexicographically ordered $(dx_1 \wedge dx_2,\, dx_1 \wedge dx_3,\, dx_2 \wedge dx_3)$: cyclic ordering ensures all three terms contribute with the same sign to the volume form.
Summing:
\begin{align*}
d\beta_F = \left( \frac{\partial F_1}{\partial x_1} + \frac{\partial F_2}{\partial x_2} + \frac{\partial F_3}{\partial x_3} \right) dx_1 \wedge dx_2 \wedge dx_3.
\end{align*}
The coefficient is the divergence $\nabla \cdot F = \sum_i \partial_{x_i} F_i$, so $d\beta_F = (\nabla \cdot F)\, dx_1 \wedge dx_2 \wedge dx_3$.
[/guided]
[/step]
[step:Derive $\nabla \times \nabla f = 0$ from $d^2 = 0$]
By the [Exterior Derivative](/theorems/1525), the operator $d$ satisfies $d^2 = 0$ on smooth differential forms of every degree. Applying this to $f \in \Omega^0(U) = C^\infty(U)$, we obtain $d(df) = 0$. By the gradient identification of the first step, $df = \alpha_{\nabla f}$, and by the curl identification of the second step applied to the vector field $\nabla f$,
\begin{align*}
0 = d(df) = d\alpha_{\nabla f} = \beta_{\nabla \times \nabla f}.
\end{align*}
Reading off coefficients in the basis $(dx_2 \wedge dx_3,\, dx_3 \wedge dx_1,\, dx_1 \wedge dx_2)$, which is linearly independent in $\Omega^2(U)$ pointwise, we conclude $(\nabla \times \nabla f)_i = 0$ for $i = 1, 2, 3$, hence $\nabla \times \nabla f = 0$ on $U$.
[/step]
[step:Derive $\nabla \cdot (\nabla \times F) = 0$ from $d^2 = 0$]
Applying $d^2 = 0$ from the [Exterior Derivative](/theorems/1525) to the $1$-form $\alpha_F \in \Omega^1(U)$, we obtain $d(d\alpha_F) = 0$. By the curl identification, $d\alpha_F = \beta_{\nabla \times F}$, and by the divergence identification applied to the vector field $\nabla \times F$,
\begin{align*}
0 = d(d\alpha_F) = d\beta_{\nabla \times F} = (\nabla \cdot (\nabla \times F))\, dx_1 \wedge dx_2 \wedge dx_3.
\end{align*}
Since $dx_1 \wedge dx_2 \wedge dx_3$ is nowhere zero on $U$, we conclude $\nabla \cdot (\nabla \times F) = 0$ on $U$. This completes the proof of all three identifications and both classical identities.
[/step]
