Hodge Laplacian and scalar

I'm reading Nakahara GEOMETRY, TOPOLOGY AND PHYSICS now.Then, Hodge Laplacian is given by \begin{align} \Delta=(d+d^{\dagger})^2=dd^{\dagger}+d^{\dagger}d \end{align} For example, we consider 0-form $$f$$, then $$d^{\dagger}f=0$$ \begin{align} \Delta f&=d^{\dagger}df=d^{\dagger}(\partial_{\mu}f dx^{\mu})\\ &=-\ast d \ast(\partial_{\mu}f dx^{\mu})=-\ast d(\frac{\sqrt{|g|}}{(m-1)!}\partial_{\mu}f g^{\mu \lambda}\epsilon_{\lambda \nu_2 \cdot \nu_m} dx^{\nu_2}\wedge \cdots \wedge dx^{\nu_m})\\ &=-\ast \frac{1}{(m-1)!}\partial_{\nu}[\sqrt{|g|}g^{\mu \lambda} \partial_{\mu}f] \epsilon_{\lambda \nu_2 \cdot \nu_m} dx^{\nu}\wedge dx^{\nu_2}\wedge \cdots \wedge dx^{\nu_m} \end{align} So far, so good, but I don't understand the next transformation. \begin{align} \Delta f&=-\ast \partial_{\nu}[\sqrt{|g|}g^{\mu \lambda} \partial_{\mu}f] g^{-1} dx^1\wedge \cdots \wedge dx^m\\ &=-\frac{1}{\sqrt{|g|}}\partial_{\nu}[\sqrt{|g|}g^{\mu \lambda} \partial_{\mu}f] \end{align} Why did $$g^{-1}$$ come up? I would appreciate it if you could tell me. For your information, $$\ast$$ is defined as follows. \begin{align} \ast (dx^{\mu_1}\wedge dx^{\mu_2}\wedge \cdots \wedge dx^{\mu_r})=\frac{\sqrt{|g|}}{(m-r)!}\epsilon^{\mu_1\mu_2 \cdots \mu_r}\ _{\nu_{r+1}\cdots \nu_{m}}dx^{\nu_{r+1}}\wedge \cdots \wedge dx^{\nu_{m}} \end{align} where $$m$$ is just dimension of manifold.

$$-\ast \frac{1}{(m-1)!}\partial_{\nu}[\sqrt{|g|}g^{\mu \lambda} \partial_{\mu}f] \epsilon_{\lambda \nu_2 \cdots\nu_m} dx^{\nu}\wedge dx^{\nu_2}\wedge \cdots \wedge dx^{\nu_m}.$$
$$$$\ast dx^{\nu}\wedge dx^{\nu_2}\wedge \cdots \wedge dx^{\nu_m}=\sqrt{|g|}\epsilon^{\nu\nu_2\cdots\nu_m}.$$$$
$$$$\frac{1}{(m-1)!}\partial_{\nu}[\sqrt{|g|}g^{\mu \lambda} \partial_{\mu}f] \epsilon_{\lambda \nu_2 \cdots\nu_m}\sqrt{|g|}\epsilon^{\nu\nu_2\cdots\nu_m}$$$$
Using equation (7.171b) of the book, we can express the Levi Civita tensor with all contravariant indices in terms of the tensor with all covariant indices, $$$$\frac{1}{(m-1)!}\partial_{\nu}[\sqrt{|g|}g^{\mu \lambda} \partial_{\mu}f] \epsilon_{\lambda \nu_2 \cdots\nu_m}\sqrt{|g|}g^{-1}\epsilon_{\nu\nu_2\cdots\nu_m}=\frac{1}{(m-1)!}\partial_{\nu}[\sqrt{|g|}g^{\mu \lambda} \partial_{\mu}f] \frac{1}{\sqrt{|g|}}(m-1)!\delta_{\lambda}^{\nu}.$$$$ The second expression follows from a property of the Levi Civita tensor (see Wikipedia). This gives the desired expression.