Let $X_1, X_2, \dots, X_n$ be i.i.d. Pareto random variables with shape parameter equal to $1$ and scale parameter $\nu$, in other words, they have p.d.f. : $$f_{X_i} (x_i) = \frac{\nu}{x^2} I_{[\nu, \infty)}(x) .$$ I need to prove that $2T$ has Chi-squared distribution, where: $$ T= \log\left[ \frac{\prod_{i=1}^n X_i }{ \min(X_i)^n} \right] .$$ What I want to prove is that $T$ has Gamma distribution, $\Gamma(n-1,1)$ and this implies that $2T$ has distribution $\chi^2(n-1)$.
This idea comes from computing the distribution of $Y_i := \log(X_i)$. With the change of variable theorem, I have that $Y_i$ has p.d.f.: $$f_{Y_i}(y_i) = \nu e^{-y_i} I_{[\log(\nu), \infty)}(y_i), $$ which is almost an exponential distribution.
We also have: $$ T = \sum_{i=1}^{n} Y_i - log(min(X_i)),$$ there is one element in the sum that is equal to zero (the minimum). If I prove that the other $n-1$ summands, $Y_i - log(min(X_i))$, are exponential(1) I finish. Any idea of how to prove it?