Two random variables $U$, $V$ such that the conditional distribution of $U$ given $V = v$ is $ve^{-u}$ for $u > ln(v)$. Prove the following:

Question

Given two random variables $U$ and $V$ such that the conditional PDF of $U$ given $V = v$ is $ve^{-u}$ for $u > ln(v)$, prove $U - V$ conditional on $V = v$ follows $Exp(1)$.

Intuitively, why is this true? My attempt:

Let $Z = U - V$.

$P(Z <= z | V = v) = P(U - v <= z | V = v) = P(U <= z + v | V = v)$

Differentiating with respect to $z$, we see that the desired PDF is $ve^{-(z+v)}$. But this is not an exponential distribution with parameter $1$.. what have I done wrong?