How to prove that $(\tau^v)^* = \tau^{-v}$ for distributions?

Question

I remember reading once that the translation operator $\tau^v f := f(\cdot + v)$ satisfies $(\tau^v)^* = \tau^{-v}$ in the sense of distributions. I believe $*$ operator refers to the adjoint, and in this case it makes sense for e.g. $f,g \in L^2$ since \begin{equation*} \int \tau^v f(x) g(x) dx = \int f(x-v) g(x) dx = \int f(z) g(z+v) dz = \int f(z) \tau^{-v} g(z) dz, \end{equation*} but how would one prove it for a general distributional $f$?