= Solution
Represent a possibly randomized <statistical test> by a function $\varphi(Y)$ taking values in $[0,1]$, its conditional probability of rejection. Its <power function> is $\pi(\rho)=\mathbb E_\rho\varphi(Y)$, and its <size of a statistical test> is the supremum of this probability over the <null hypothesis>. At a prescribed <significance level> $\alpha$, a <uniformly most powerful test> has at least as much power as every other level-$\alpha$ test at every parameter in the alternative.
Put $S=\sum_{j=1}^nY_j$. For any fixed alternative $0<\rho_1<\rho_0$, the <likelihood ratio> is
$$
\frac{L(\rho_1)}{L(\rho_0)}=\left(\frac{\rho_1}{\rho_0}\right)^n e^{(\rho_0-\rho_1)S},
$$
which is strictly increasing in $S$. The <Neyman-Pearson lemma> therefore selects the upper tail $S>c_\alpha$. Its cutoff is determined solely by the null distribution, not by $\rho_1$, so the same <statistical test> is most powerful against every allowed alternative and hence uniformly most powerful.
Under rate $\rho$, $S$ has a <gamma distribution> with shape $n$ and rate $\rho$. For completeness, convolution starts with $\rho e^{-\rho s}$ and, if the $n$-term density is $\rho^ns^{n-1}e^{-\rho s}/(n-1)!$, convolving once more gives $\rho^{n+1}s^ne^{-\rho s}/n!$. Integrating this density by parts repeatedly gives its upper-tail probability. Consequently \b[reject precisely when $S>c_\alpha$], where
$$
e^{-\rho_0c_\alpha}\sum_{j=0}^{n-1}\frac{(\rho_0c_\alpha)^j}{j!}=\alpha,
$$
and the <power function> is
$$
\boxed{\pi(\rho)=e^{-\rho c_\alpha}\sum_{j=0}^{n-1}\frac{(\rho c_\alpha)^j}{j!}}.
$$
For $0<\alpha<1$ the continuous strictly decreasing null tail gives a unique positive cutoff, with exact size $\alpha$ and no boundary randomization.
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