= Solution
Rearranging the outgoing <radial null geodesics of the Vaidya metric> equation along $r=r_0(v)$ gives
$$
\boxed{M(v)=\frac{r_0(v)}2\left(1-2\frac{dr_0}{dv}\right).}
$$
For the prescribed parametrization by $w$, the <chain rule> yields
$$
\frac{dv}{dw}=1+2\beta n w^{n-1}>0,
\qquad
\frac{dr_0}{dv}=\frac{\beta n w^{n-1}}{1+2\beta n w^{n-1}},
\qquad
M(v(w))=\frac{\beta w^n}{2(1+2\beta n w^{n-1})}.
$$
Since $n>1$, $v/w=1+2\beta w^{n-1}\to1$ as $w\to0^+$. The <asymptotic equivalence> $w\sim v$ therefore implies
$$
\boxed{M(v)\sim\frac\beta2 v^n,\qquad c_1=\frac\beta2.}
$$
More precisely $w=v+O(v^n)$ and $M(v)=(\beta/2)v^n[1+O(v^{n-1})]$, using <Big O notation>. The positivity of $dv/dw$ justifies treating $w$ as a single-valued function of $v$ locally.
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