= Solution
After a record of value $v$, the first subsequent observation exceeding it has the original distribution conditioned on $X>v$. Indeed, summing over the number of failed observations gives the density
$$
\sum_{j=0}^\infty F(v)^j f(w)\mathbf1_{w>v}=\frac{f(w)}{1-F(v)}\mathbf1_{w>v}.
$$
Independence makes this conditional transition depend only on the latest record. Multiplying these transitions, starting with the density $f(v_1)$, proves
$$
\boxed{f_{V_1,\ldots,V_n}=\mathbf1_{0<v_1<\cdots<v_n}\,f(v_1)\prod_{j=2}^n\frac{f(v_j)}{1-F(v_{j-1})}.}
$$
The ordering indicator is part of the density; it is zero elsewhere. Continuity of the observation distribution avoids ties.
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