Consider an exponential model with density $\theta e^{-\theta x}$ with $x> 0$ and $\theta >0 $. Derive LR test of approximate level $\alpha$ (For large sample size) for the hypothesis problem
$H_0: \theta = \theta_0$ Versus $H_1:\theta \leq \theta_0$
Based on iid sample $X_1,...,X_n$
SOLUTION:
$$ \Lambda = \frac{\theta^n e^{-\theta_0 \sum X_i}}{\bar{X}^{-n}e^{-\sum {\frac{X_I}{\bar{X}}}}} = (\theta_0\bar{X})^ne^{-n\theta_0\bar{X}+n}$$
I do not understand the last simplification
EDIT (after @callculus answer)
Level $\alpha$:
$$ 2 \Lambda log = 2n {\theta_0\bar{X}-1-log(\theta_0\bar{X})}$$
For large size $n$ The rejection region is given by $2n {\theta_0\bar{X}-1-log(\theta_0\bar{X})} > \chi^2_1$. $\chi^2_1$ Denoting the $(1-\alpha)100%$ percentile of the $\chi^2_1$ distibution.
How can i see that the distribution is $\chi^2_1?$
Firstly we should have $\frac{\theta_{\color{red}0}^n}{\overline X^{-n}}=\theta_{0}^n\cdot \overline X^{n}=\left(\theta_{0}\overline X\right)^n$
Similar for the next fraction.
$$\Large{\frac{ e^{-\theta_0 \sum\limits_{i=1}^n X_i}}{e^{-\sum\limits_{i=1}^n {\frac{X_i}{\bar{X}}}}}}\normalsize= e^{-\theta_0 \sum\limits_{i=1}^n X_i+\sum\limits_{i=1}^n {\frac{X_i}{\bar{X}}}}$$
Let´s focus on the exponent.
$\frac1n\sum\limits_{i=1}^n X_i=\overline X\Rightarrow \sum\limits_{i=1}^n X_i=n\cdot \overline X$
Therefore $\sum\limits_{i=1}^n {\frac{X_i}{\bar{X}}}=\frac1{\overline X}\cdot \sum\limits_{i=1}^n X_i=n$
Consequently
$-\theta_0 \sum\limits_{i=1}^n X_i+\sum\limits_{i=1}^n {\frac{X_i}{\bar{X}}}=-\theta_0\cdot n\cdot \overline X+n$