Volume of a solid in n dimensions. I'm learning about the volume of a sphere in $n$ dimensions. All the proofs I've found have started with the assumption that the volume of an $n$-sphere of radius $R$ is $V_nR^n$, where $V_n$ is the volume of the unit sphere. What is the actual proof for this? I've just seen it handwaved as an obvious fact of dimensional analysis.
To be clear, I'm defining an n-sphere of radius $R$ to be the set of points such that $x_1^2 + x_2^2 + \cdots + x_n^2 \le R^2$
 A: "Volume" in Euclidean space is another way of saying Lebesgue measure, and $n$-sphere should be called $n$-ball instead if you want to talk about its volume.
Some notation: $B$ is the unit ball in $\mathbb R^n$. $\lambda^n$ is the Lebesgue measure in $\mathbb R^n$. $1_S$ is the characteristic function of $S$.
The $n$-ball of radius $r$ is simply $\Phi[B]$, where $\Phi$ is a linear transformation given by the diagonal matrix
$$ \Phi = \begin{bmatrix} r & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & r \end{bmatrix}
$$
We can factor this matrix into a series of elementary matrices, with $\Phi = \prod_{k=0}^{n-1} \Phi_k$, where $\Phi_k$ is the diagonal matrix with $k$th element on the diagonal set to $r$. Using Fubini's Theorem, we have
$$ \lambda^n(\Phi[B]) = \int 1_{\Phi[B]} \mathrm d\lambda^n = \idotsint_S 1_{\Phi[B]} \mathrm d\lambda \cdots \mathrm d\lambda
$$
and we can apply the identity to each $\lambda$:
$$ \int f(x)\;\mathrm dx = |r| \int f(rx)\;\mathrm dx
$$
Hence
$$ \lambda^n(\Phi[B]) = \idotsint 1_{\Phi[B]} \mathrm d\lambda \cdots \mathrm d\lambda = |r|^n \idotsint 1_B \mathrm d\lambda \cdots \mathrm d\lambda = |r|^n \lambda^n(B)
$$
