= Solution
A <critical region> $C$ is the set of sample outcomes on which a test rejects its <null hypothesis>. Its <power function> is $\beta(\theta)=\mathbb P_\theta(X\in C)$; its <size of a statistical test> is $\sup_{\theta\in\Theta_0}\beta(\theta)$. Randomized tests replace the indicator of $C$ by a rejection <probability> $\phi(X)\in[0,1]$.
The <Neyman-Pearson lemma> states that for two simple hypotheses with densities $p_0,p_1$, the test $\phi^*$ that is one where $p_1>kp_0$, zero where $p_1<kp_0$, and randomized on equality to have null rejection <probability> $\alpha$, maximizes power among tests of size at most $\alpha$. Here $k\ge0$. To prove it, let $\phi$ be any competing test. Pointwise, $(\phi^*-\phi)(p_1-kp_0)\ge0$, since both factors have the same sign off the equality set. Integrating gives
$$
\mathbb E_1\phi^*-\mathbb E_1\phi\ge k(\mathbb E_0\phi^*-\mathbb E_0\phi)\ge0.
$$
This proves the lemma, including equality-set randomization and points where $p_0=0$.
Take the exponential parameter to be its rate: $p_\lambda(x)=\lambda e^{-\lambda x}$ for $x\ge0$. For $T=\sum_iX_i$, the joint <likelihood ratio> is $(\lambda_1/\lambda_0)^n e^{-(\lambda_1-\lambda_0)T}$, decreasing in $T$. Hence for $0<\alpha<1$ the most powerful test is
$$
\boxed{\text{reject when }T\le c_\alpha,\qquad \mathbb P_{\lambda_0}(T\le c_\alpha)=\alpha.}
$$
There is no boundary randomization because $T$ has a continuous <gamma distribution>. Its density follows by repeated convolution: $p_T(t)=\lambda^nt^{n-1}e^{-\lambda t}/(n-1)!$. Integrating by parts gives the <exponential-rate likelihood-ratio test> power
$$
\boxed{\beta(\lambda)=1-e^{-\lambda c_\alpha}\sum_{j=0}^{n-1}\frac{(\lambda c_\alpha)^j}{j!}.}
$$
The threshold is the unique positive solution of this formula at $\lambda=\lambda_0$ with value $\alpha$. Differentiating the finite sum yields
$$
\beta'(\lambda)=c_\alpha e^{-\lambda c_\alpha}\frac{(\lambda c_\alpha)^{n-1}}{(n-1)!}>0.
$$
Thus the <power function> is strictly increasing, and its supremum over $0<\lambda\le\lambda_0$ is $\beta(\lambda_0)=\alpha$. Therefore \b[the same test has size $\alpha$ for the composite null $\lambda\le\lambda_0$]. Since the threshold does not depend on the particular larger rate, it is also uniformly most powerful against all $\lambda>\lambda_0$. The endpoint sizes zero and one have their trivial constant tests.
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