Suppose that $X_1, \ldots, X_n$ come from the distribution
$$ f(x|\theta) = e^{-(x-\theta)} \ \ \text{for} \ \ x \geq \theta \ \ \text{and} \ \ 0 \ \ \text{otherwise} $$
Define $X_{(1)} = \min_{1 \leq i \leq n}X_i$ as the smallest value among the $X_i$.
I want to find the distribution of the minimum of $X_i$ under the constrained parameter space $-\infty < \theta\leq \theta_0$ where $\theta_0$ is some constant. Normally, this is taken as $-\infty < \theta\leq \infty$.
In my book, it states that:
$$ P(X_{(1)} \geq c) = e^{-n(c-\theta_0)} $$
However, I cannot find this answer on my own. It seems that I am always left with $e^{-n(c-\theta)}$ instead and don't know how to incorporate the $\theta_0$. Does anyone have any ideas or tips?