Hypothesis testing: find the UMP test

Question

Suppose $X_1,\dots,X_n$ are i.i.d. They are distributed as follows: $P(X_i > x)=(1+x)^{-\lambda}$ where $x\geq 0$ and $\lambda> 0$. I have to test the following hypothesis with level $\alpha_0$; $H_0:\lambda\leq \lambda_0$ versus $H_1:\lambda>\lambda_0$. First I looked at a the hypothesis $H'_0:\lambda=\lambda_0$ versus $H_1:\lambda>\lambda_0$. After using the Neyman-Pearson lemma, I concluded that $\sum_{i=1}^n \log (X_i+1)$ must be small, so the critical region becomes $K=\{x\in\mathbb{R}^n:\sum_{i=1}^n \log (X_i+1)\leq c\}$ for some constant $c$ which can be found by $P_{\lambda_0}(X\in K)=\alpha_0$ where $X=(X_1,\dots,X_n)$. By the Neyman-Pearson lemma this test is UMP for testing $H'_0:\lambda=\lambda_0$ versus $H_1:\lambda>\lambda_0$. Now I want to show that this is also UMP for the test $H_0:\lambda\leq \lambda_0$ versus $H_1:\lambda>\lambda_0$, which means I have to show that for $\lambda\leq \lambda_0$: $P_{\lambda}(X\in K)\leq \alpha_0$, but I don't know how to do this. There is a hint which says $P_{\lambda}(\lambda \log (X_1+1)\leq z)=P_{\lambda_0}(\lambda_0 \log (X_1+1)\leq z)$ for $z\geq 0$, which can be easily shown. I really need help. Thanks.