Let $(M, g)$ be an oriented Riemannian manifold with volume form $dv$. Let $*$ be the Hodge star operator defined by the relation $\omega \wedge *\tau = g(\omega, \tau) dv$ where $g$ is extended to forms. I know that w.r.t. the global product given by the extension of $g$ to forms, $\delta:= *d*$ is the adjoint of $d$ up to a sign. That is $\int_M g(d\omega, \tau)dv = \int_M g(\omega, \delta \tau)dv$.
I am trying to see that given $\tau \in \Omega^1(M)$, we have that $(\delta \tau) dv = d(*\tau)$ up to a sign. I tried using what I know and got at $$ (\delta \tau ) dv= ( * d * \tau ) dv = dv \wedge ( *d* \tau) = g(dv, d* \tau) dv = g(\delta dv, *\tau) dv = (\delta dv) \wedge (** \tau) $$
So, since $** $ is the identity with a sign, $(\delta \tau) dv = (\delta dv) \wedge \tau$ up to sign. This doesn't seem to help, though.
It suffices to use $*1 = dv$, $** = \pm 1$ and $g(\omega, \tau) = g(*\omega, *\tau)$.
\begin{align} \delta \tau dv &= (\pm *d* \tau) dv\\ &= \pm g(1,* d*\tau)dv \\ &= \pm g(dv, d(*\tau))dv \\ &= \pm g(d(*\tau), dv)dv \\ &= \pm d(*\tau)\wedge *dv \\ &= \pm d(*\tau) \end{align}