Given $T=(X_{(1)},\sum[X_i-X_{(1)}])$ is minimum sufficient statistics for exp(a,b).

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Given $T=(X_{(1)},\sum[X_i-X_{(1)}])$ is minimum sufficient statistics for exp(a,b). Show that $X_{(1)}$ and $\sum[X_i-X_{(1)}]$ are independently distributed as $Exp(a,b/n)$ and $\frac{1}{2}b\chi_{(n-2)}^2$, respectively.

I have supposed a new variable $Y_1=X_{(1)}$, $Y_i=X_{(i)}-X_{(i-1)}$, i=2,3,... and I have found the pdf of $Y_i$, and then I would like to show each $Y_i$ independently first, but I failed...

Do you have any other ideas for this question? I am struggling now.

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I am know confused that why $\sum[X_i-X_{(1)}]$ could follows chi-squared distribution, since I found Y_i is exponential and $X_i-X_{(1)}$ is just gamma, how could I figure out its chi-squared?