Why $m(B^n(0, r)) = c_nr^n$? (Bear with me: I realize this is quite basic question, but I'm a little loss at  how to search for an answer). Anyway: 
I came across a real analysis proof which uses a property that (as far as I can see) amounts to this:
Suppose $B^n(0, r)$ is a $n$-dimensional (open) ball (in $\mathbb{R}^n$) and $m$ is a (complete) measure.
Now $m(B^n(0, r))$ can be written as $c_nr^n$, $c_n$ being a constant dependent only on dimension $n$.
My "gut feeling" is that this follows from translation invariance of a measure (is this correct?), but the result isn't immediate to me.

Indeed it does. The translation invariance makes it into a Haar measure, and it's easily verified that it holds for Lebesgue measure (standard integration),
$$\int_{B(0,r^n)}\,dV=r^n\int_{B(0,1)}dV'=c_nr^n$$
holds by change of coordinates, with $c_0$ being the Lebesgue measure of the unit ball, which we see by letting
$$\begin{cases} y_i=rx_i \\dy_i=rdx_i \\dV=r^ndV'\end{cases}$$
