Derive formula for volume element of manifold

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In a course regarding analysis of manifolds, I was given the following question:

$M\subseteq\mathbb{R}^n$ is a smooth $(n-1)$-dimensional manifold, which is described by a graph of a function $f:U\rightarrow\mathbb{R}$ where $U\subseteq\mathbb{R}^{n-1}$ is an open set and $r:U\rightarrow M$ is the parameterization given by $r\left(x_{1},...,x_{n-1}\right)=\left(x_{1},...,x_{n-1},f\left(x_{1},...,x_{n-1}\right)\right)$ . Then the volume of the manifold is:

$$\intop_{M}1dV=\intop_{U}\sqrt{1+\Vert\nabla f\Vert^{2}}du_{1}du_{2}...du_{n-1}$$

I've tried proving this using induction on the the dimension of the manifold but I get stuck in some real gritty algebra.

I know how to prove this for the case $n=3$, where the volume element is the square root of the Gram matrix made from each partial derivative.

$$\det\left(\mathrm{Gram}\left(\frac{\partial r}{\partial x_{1}},\frac{\partial r}{\partial x_{2}}\right)\right) =\left\langle \frac{\partial r}{\partial x_{1}},\frac{\partial r}{\partial x_{1}}\right\rangle \cdot\left\langle \frac{\partial r}{\partial x_{2}},\frac{\partial r}{\partial x_{2}}\right\rangle -\left\langle \frac{\partial r}{\partial x_{1}},\frac{\partial r}{\partial x_{2}}\right\rangle \cdot\left\langle \frac{\partial r}{\partial x_{2}},\frac{\partial r}{\partial x_{1}}\right\rangle =\left(1+\left|\frac{\partial f}{\partial x_{1}}\right|^{2}\right)\cdot\left(1+\left|\frac{\partial f}{\partial x_{2}}\right|^{2}\right)-\left(\left|\frac{\partial f}{\partial x_{1}}\right|\left|\frac{\partial f}{\partial x_{2}}\right|\right)^{2} =1+\left|\frac{\partial f}{\partial x_{1}}\right|^{2}+\left|\frac{\partial f}{\partial x_{2}}\right|^{2}=1+\Vert\nabla f\Vert^{2}$$

So the volume element comes out the way I want it. But the next transition is the one I get stuck on:

$$\det\left(\mathrm{Gram}\left(\frac{\partial r}{\partial x_{1}},...,\frac{\partial r}{\partial x_{n-1}}\right)\right)=1+\Vert\nabla f\Vert^{2}$$ I would appreciate a nudge in the right direction.

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First we notice $$\mathrm{Gram}\left(\frac{\partial r}{\partial x_{1}},...,\frac{\partial r}{\partial x_{n-1}}\right)= I_{n-1} + \nabla f\cdot(\nabla f)^\mathrm{T}$$ Choose an orthogonal matrix $O$ such that $O(\nabla f) = (\Vert\nabla f\Vert,0,\dots,0)^\mathrm{T}$, then $$\det\left(\mathrm{Gram}\left(\frac{\partial r}{\partial x_{1}},...,\frac{\partial r}{\partial x_{n-1}}\right)\right) = \det\left(I_{n-1} +\nabla f\cdot(\nabla f)^\mathrm{T}\right) = \det\left(OO^\mathrm{T} +O\nabla f\cdot(\nabla f)^\mathrm{T} O^\mathrm{T}\right) = \det\left(\text{diag}(1+\Vert\nabla f\Vert^2,1,\dots,1)\right) = 1+\Vert\nabla f\Vert^2$$