$M \subset \mathbb{R}^{N}$ is a (oriented) $n-1$ dimensional submanifold. Suppose $\nu \in T_{p}M^{\bot}$, of length one (a normal unit vector on $M$).
How and why does the contraction $\nu_{\neg}(dx_{1}\wedge...\wedge dx_{n})$ (yes the contraction symbol is reversed, sorry) give the canonical volume form, vol$_{M}$ (or vol$_{g}$ ) , on $M$?
There is a theorem that says that the vol$_{g}(p)$ can be written as $\sqrt{\det g_{ij }}dy_{1}\wedge ... \wedge dy_{n-1}$, where $y_i$ would be local coordinates of $M$. Do I need to do something with this. Or something with the fact that vol$_{g}(X_1,...,X_{n-1}) = 1$ for an oriented orthonormal basis $\{X_1,...X_{n-1}\}$ of $T_{p}M$. I'm kind of getting lost in what I can and have to use.
(The contraction was defined as $\nu_{\neg}\alpha(v_1,...,v_{n-1}) = \alpha(\nu \wedge v_1 \wedge , ... , \wedge v_{n-1})$, where $\alpha$ is an n-form.
At every point $p\in M$, you can extend an (oriented) orthonormal basis $v_1,\dots,v_{n-1}$ of $T_pM$ to an (oriented) orthonormal basis $\nu,v_1,\dots,v_{n-1}$ of $T_p\mathbb{R}^n$. So $$\mathrm{d}vol_M(p)=v_1^\flat\wedge\dots\wedge v_{n-1}^\flat=\iota_\nu(\nu^\flat\wedge v_1^\flat\wedge\dots\wedge v_{n-1}^\flat)=\iota_\nu\mathrm{d}vol_{\mathbb{R}^n}.$$