The Setup
Let there be a volume differential form
\begin{equation} dV=\frac{1}{d!}\epsilon_{\mu_1\dots\mu_d}dx^{\mu_1}\wedge\dots\wedge dx^{\mu_d} \end{equation}
on a d-dimensional manifold $M$ with coordinates $x^\mu$. We can build the differential form of a hypersurface $S\subset M$ with dimension $d-n$ using $n$ normal vectors $N^\mu_i$ with $i=1,\dots,n$
\begin{equation} dS={N_1}^{\mu_1}\dots {N_n}^{\mu_n} d\Sigma_{\mu_1\dots\mu_n} \end{equation}
where we have defined
\begin{equation} d\Sigma_{\mu_1\dots\mu_n}=\frac{1}{(d-n)!}\epsilon_{\mu_1\dots\mu_d}dx^{\mu_{n+1}}\wedge\dots\wedge dx^{\mu_d} \end{equation}
I'm trying to prove how $d\Sigma_{\mu_1\dots\mu_n}$ transforms under a coordinate change $y^\mu =\phi^\mu(x)$.
A result I do know how to get
For the full volume $dV$ I think I know how to do it:
$$\begin{align} dV(x)&=\frac{1}{d!}\epsilon_{\mu_1\dots\mu_d}dx^{\mu_1}\wedge\dots\wedge dx^{\mu_d}\\\\ &=\frac{1}{d!}\epsilon_{\mu_1\dots\mu_d}\frac{\partial x^{\mu_1}}{\partial y^{\nu_1}}\dots\frac{\partial x^{\mu_d}}{\partial y^{\mu_d}}dy^{\mu_1}\wedge\dots\wedge dy^{\mu_d}\\\\ &=\frac{1}{d!}\epsilon_{\mu_1\dots\mu_d}\frac{\partial x^{\mu_1}}{\partial y^{\nu_1}}\dots\frac{\partial x^{\mu_d}}{\partial y^{\mu_d}}\delta ^{\mu_1} _{[\rho_1}\dots \delta^{\mu_d} _{\rho_d]} dy^{\rho_1}\wedge\dots\wedge dy^{\rho_d}\\\\ &=\frac{1}{d!}\epsilon _{\mu_1\dots\mu_d}\frac{\partial x^{\mu_1}}{\partial y^{\nu_1}}\dots\frac{\partial x^{\mu_d}}{\partial y^{\mu_d}}\frac{1}{d!}\epsilon ^{\nu_1\dots\nu_d}\epsilon _{\rho_1\dots\rho_d} dy ^{\rho_1}\wedge\dots\wedge dy ^{\rho_d}\\\\ &=\frac{1}{d!}J \epsilon _{\rho_1\dots\rho_d} dy ^{\rho_1}\wedge\dots\wedge dy ^{\rho_d} \end{align}$$
where I defined the determinant of the Jacobian as
$$J=\epsilon_{\mu_1\dots\mu_d}\epsilon^{\nu_1\dots\nu_d}\frac{\partial x^{\mu_1}}{\partial y^{\nu_1}}\dots\frac{\partial x^{\mu_d}}{\partial y^{\mu_d}}$$
and used the identities $$\delta ^{\mu_1} _{[\rho_1}\dots\delta ^{\mu_d} _{\rho_d]}=\frac{1}{d!}\epsilon ^{\nu_1\dots\nu_d}\epsilon _{\rho_1\dots\rho_d}$$
and $$\epsilon_{\mu_1\dots\mu_d}\epsilon^{\mu_1 \dots \mu_d}=d!$$
My problem
My problem is trying to do the same calculation for $d\Sigma_{\mu_1 \dots \mu_{d-n}}$, I think the result should be
$$ d\Sigma _{\mu _1 \dots \mu _n}(x)=J \frac{\partial y ^{\nu_1}}{\partial x ^{\mu_1}}\dots \frac{\partial y ^{\nu_n}}{\partial x ^{\mu_n}} d\Sigma _{\nu _1 \dots \nu _n}(y) $$
But I'm not sure how to get there, the step where I convert the antisymmetrized Kronecker deltas into a product of two Levi-Civita symbols doesn't work since the indices don't cover the full manifold, they only go from 1 to $d-n$.
To clarify, I understand why that result should hold, I'm trying to understand how the calculation is explicitely carried out. Thanks!