This is similar to this question here: MP test construction for shifted exponential distribution.
Joint pdf of $X_1,\ldots,X_n$ is $$f_{\lambda}(x_1,\ldots,x_n)=\lambda^n\left(\prod_{i=1}^n x_i\right)^{-2}\mathbf1_{x_{(1)}>\lambda}$$
So a sufficient statistic is $X_{(1)}=\min\limits_{1\le i\le n} X_i$ rather than $\prod\limits_{i=1}^n X_i$.
For $\lambda_1>\lambda_0$, we have the ratio
\begin{align}
\Lambda(x_1,\ldots,x_n)&=\frac{f_{H_1}(x_1,\ldots,x_n)}{f_{H_0}(x_1,\ldots,x_n)}
\\\\&=\begin{cases}\left(\frac{\lambda_1}{\lambda_0}\right)^n&,\text{ if }x_{(1)}>\lambda_1 \\ \quad0&,\text{ if }\lambda_0<x_{(1)}\le \lambda_1\end{cases}
\end{align}
By Neyman-Pearson lemma, a most powerful test would reject $H_0$ for large values of $\Lambda$. Since $\Lambda$ is a non-decreasing function of $x_{(1)}$, the critical region of the test is of the form $x_{(1)}>c$. From the size/level restriction $P_{H_0}(X_{(1)}>c)=\alpha$, we get $c=\lambda_0\alpha^{-1/n}$. Hence a most powerful level $\alpha$ test for testing $H_0:\lambda=\lambda_0$ against $H_1:\lambda=\lambda_1(>\lambda_0)$ is
$$\varphi(x_1,\ldots,x_n)=\begin{cases}1&,\text{ if }x_{(1)}>\lambda_0\alpha^{-1/n} \\ 0&,\text{ otherwise }\end{cases}$$
This test is in fact uniformly most powerful for testing $H_0$ against $H_1':\lambda>\lambda_0$ because it does not depend on the choice of $\lambda_1$.