"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
$$ \lambda^n(\Phi[B]) = \idotsint 1_{\Phi[B]} \mathrm d\lambda \cdots \mathrm d\lambda = |r|^n \idotsint 1_B \mathrm d\lambda \cdots \mathrm d\lambda = |r|^n \lambda^n(B)
$$