Basu's theorem to show independence

Using Basu's theorem, prove that $\sum\limits_{i = 1 }^n {(X_i - X_{(1)}) }$ and $X_{(1)}$ are independent for any $(\theta, \lambda)$. You may assume that $X_{(1)}$ is complete and sufficient for $θ$ for any fixed $λ$,

Since $X_{(1)}$ is complete and sufficient, it is independent of every ancillary statistics. For $\sum\limits_{i = 1 }^n {(X_i - X_{(1)}) }$, I'm not completely sure how to prove it is independent. I thought about getting the joint pdf of $(X_{(1)},X_{(2)},...,X_{(n)})$ but I was not able to identify any distributions and prove that it doesn't depend on $\theta$

You can rewrite $\sum\limits_{i=1}^n (X_i-X_{(1)})$ as $$\sum_{i=1}^n (X_i-X_{(1)})=\sum_{i=1}^n\left[ (X_i-\theta)-(X_{(1)}-\theta)\right]$$

Now argue that distributions of both $X_i-\theta$ and $X_{(1)}-\theta$ are independent of $\theta$ for fixed $\lambda$, so that $\sum\limits_{i=1}^n (X_i-X_{(1)})$ is ancillary for $\theta$ for a given $\lambda$.

Since $X_{(1)}$ is complete sufficient for $\theta$, by Basu's theorem it follows that $\sum\limits_{i=1}^n (X_i-X_{(1)})$ is independent of $X_{(1)}$ for any fixed $\lambda$. Since $\lambda$ is arbitrary, your desired result follows.