Joint density of the sample $(X_1,X_2,\ldots,X_n)$ is
$$f_{\theta}(x_1,\ldots,x_n)=\exp\left(-\sum_{i=1}^n(x_i-\theta)\right)\mathbf1_{x_{(1)}>\theta}\quad,\,\theta>0$$
By N-P lemma, a most powerful test of size $\alpha$ for testing $H_0:\theta=\theta_0$ against $H_1:\theta=\theta_1(>\theta_0)$ is given by $$\varphi(x_1,\ldots,x_n)=\begin{cases}1&,\text{ if }\lambda(x_1,\ldots,x_n)>k\\0&,\text{ if }\lambda(x_1,\ldots,x_n)<k\end{cases}$$
, where $$\lambda(x_1,\ldots,x_n)=\frac{f_{\theta_1}(x_1,\ldots,x_n)}{f_{\theta_0}(x_1,\ldots,x_n)}$$
and $k(>0)$ is such that $$E_{\theta_0}\varphi(X_1,\ldots,X_n)=\alpha$$
Now,
\begin{align}
\lambda(x_1,\ldots,x_n)&=\frac{\exp\left(-\sum_{i=1}^n(x_i-\theta_1)\right)\mathbf1_{x_{(1)}>\theta_1}}{\exp\left(-\sum_{i=1}^n(x_i-\theta_0)\right)\mathbf1_{x_{(1)}>\theta_0}}
\\\\&=e^{n(\theta_1-\theta_0)}\frac{\mathbf1_{x_{(1)}>\theta_1}}{\mathbf1_{x_{(1)}>\theta_0}}
\\\\&=\begin{cases}e^{n(\theta_1-\theta_0)}&,\text{ if }x_{(1)}>\theta_1\\0&,\text{ if }\theta_0<x_{(1)}\le \theta_1\end{cases}
\end{align}
So $\lambda(x_1,\ldots,x_n)$ is a monotone non-decreasing function of $x_{(1)}$, which means
$$\lambda(x_1,\ldots,x_n)\gtrless k \iff x_{(1)}\gtrless c$$, for some $c$ such that $$E_{\theta_0}\varphi(X_1,\ldots,X_n)=\alpha$$
We thus have
$$\varphi(x_1,\ldots,x_n)=\begin{cases}1&,\text{ if }x_{(1)}>c\\0&,\text{ if }x_{(1)}<c\end{cases}$$
Again,
\begin{align}
E_{\theta_0}\varphi(X_1,\ldots,X_n)&=P_{\theta_0}(X_{(1)}>c)
\\&=\left(P_{\theta_0}(X_1>c)\right)^n
\\&=e^{n(\theta_0-c)}\quad,\,c>\theta_0
\end{align}
So from the size condition we get $$c=\theta_0-\frac{\ln\alpha}{n}$$
Finally, the test function is
$$\varphi(x_1,\ldots,x_n)=\begin{cases}1&,\text{ if }x_{(1)}>\theta_0-\frac{\ln\alpha}{n}\\0&,\text{ if }x_{(1)}<\theta_0-\frac{\ln\alpha}{n}\end{cases}$$
