= Solution
The <exterior derivative> is a linear map $d:\Omega^k(M)\to\Omega^{k+1}(M)$, agrees with the ordinary differential on functions, satisfies $d^2=0$, and obeys the graded <Leibniz rule>
$$
d(\alpha\wedge\beta)=d\alpha\wedge\beta+(-1)^k\alpha\wedge d\beta\qquad(\alpha\in\Omega^k).
$$
In local coordinates it differentiates the coefficient functions:
$$
d\left(\frac1{k!}\alpha_{i_1\cdots i_k}dx^{i_1}\wedge\cdots\wedge dx^{i_k}\right)
=\frac1{k!}\partial_j\alpha_{i_1\cdots i_k}dx^j\wedge dx^{i_1}\wedge\cdots\wedge dx^{i_k}.
$$
Antisymmetry cancels the second partial derivatives, proving $d^2=0$ locally. These properties determine the coordinate-independent derivative.
For any smooth $\phi:N\to M$, the <pullback of a differential form> is defined by
$$
(\phi^*\alpha)_x(v_1,\ldots,v_k)
=\alpha_{\phi(x)}(d\phi_xv_1,\ldots,d\phi_xv_k).
$$
There is no requirement that $\phi$ be invertible or an immersion. Pullback preserves wedge products and sends $df$ to $d(f\circ\phi)$ by the chain rule. Locally every form is a sum of $f\,dx^{i_1}\wedge\cdots\wedge dx^{i_k}$; applying the preceding identities and $d^2\phi^i=0$ proves
$$
\boxed{d(\phi^*\alpha)=\phi^*(d\alpha).}
$$
Describe the <brane> by an embedding $X:\Sigma\to M$, with nondegenerate <induced worldvolume metric> $h=X^*g$. Locally the <Poincaré lemma> gives $F=dA$ because $F$ is a <closed differential form>. For a timelike worldvolume, the geometric area and <Wess-Zumino brane coupling> give the action
$$
\boxed{S[X]=-T\int_\Sigma d^{p+1}\sigma\,\sqrt{|\det h|}
+q\int_\Sigma X^*A,\qquad
h_{\alpha\beta}=g_{\mu\nu}(X)\partial_\alpha X^\mu\partial_\beta X^\nu.}
$$
$T$ is the <brane tension> and $q$ the charge in this normalization. The geometric term is the higher-dimensional version of the <Nambu–Goto action>. Both integrals are invariant under changes of worldvolume coordinates.
The additive gauge transformation is $A\mapsto A+d\Lambda$. Its change of action is
$$
\delta_\Lambda S=q\int_\Sigma X^*d\Lambda
=q\int_\Sigma d(X^*\Lambda)
=q\int_{\partial\Sigma}X^*\Lambda.
$$
The last step is the <Generalized Stokes theorem>. It vanishes for a closed worldvolume, or with suitable boundary restrictions. An open worldvolume can instead have a compensating boundary coupling, as in <gauge invariance of an open-brane coupling>.
More directly, for a variation vector $V$ of the embedding, <Cartan's magic formula> gives
$$
\delta\int_\Sigma X^*A
=\int_\Sigma X^*(\iota_VF)+\int_{\partial\Sigma}X^*(\iota_VA).
$$
Thus the interior force depends on $F$, not the gauge potential. Define the trace of the <second fundamental form> by
$$
K^\mu=\frac1{\sqrt{|h|}}\partial_\alpha\left(\sqrt{|h|}h^{\alpha\beta}\partial_\beta X^\mu\right)
+\Gamma^\mu_{\nu\rho}h^{\alpha\beta}\partial_\alpha X^\nu\partial_\beta X^\rho.
$$
The area variation is $-\int\sqrt{|h|}\,g(V,K)$ plus a boundary term. For compactly supported interior variations the resulting equations are
$$
\boxed{T\sqrt{|h|}\,K_\mu
+\frac{q}{(p+1)!}\varepsilon^{\alpha_0\cdots\alpha_p}
F_{\mu\nu_0\cdots\nu_p}\partial_{\alpha_0}X^{\nu_0}\cdots\partial_{\alpha_p}X^{\nu_p}=0.}
$$
They exhibit the <field-strength force on a charged brane> and are gauge invariant.
Two global qualifications are necessary. First, replacing $A$ by $d\Lambda$ alone, as printed, gives $dA'=0$ and would erase a nonzero $F$; \b[the intended transformation must be the additive shift] above. Second, closedness of $F$ only guarantees a potential locally. If a global potential does not exist and the worldvolume bounds a filling $B$, its charged term can be defined as $q\int_B F$. Different fillings differ by a period of $F$; independence of the quantum phase requires $q\int_C F\in2\pi\hbar\mathbb Z$ for the corresponding closed cycles. Without the required global potential or patching/filling data, the given closed form still defines the local variational equations, but not automatically a single globally defined real-valued action.
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