Let $\ast$ be the star operator $$\ast:\Lambda^p(V)\to \Lambda^{d-p}(V)$$ so that we have $$\ast(e_{i_1}\wedge...\wedge e_{i_p})=e_{j_1}...\wedge e_{j_{d-p}}$$ where $$e_{i_1},...,e_{i_p},e_{j_1},...,e_{j_{d-p}}$$ is a positive basis of $V$. For $v,w\in\Lambda^p(V)$ we have $$\langle v,w\rangle=\ast(w\wedge \ast v)=\ast(v\wedge \ast w).$$
Why then $$\ast(1)=\sqrt{\det(g_{ij})}dx^1\wedge ...\wedge dx^d$$ and also how do we use the above property to yield $$\int_M \langle\alpha,\beta\rangle\ast(1)=\int_M \alpha\wedge \ast\beta$$ for $\alpha,\beta\in \Omega^p(M)$ with compact support?