Gamma is given by $$f_X{(x)} = \frac{\lambda^a x^{a-1}e^{-\lambda x}}{\Gamma(a)}$$
I can remove the constant being $\frac{x^{a-1}}{\Gamma(a)}$
$$L(\theta | x_{1}, x_2, \cdots, x_n = \bar{x}) = \prod_{i=1}^{n} \theta^a e^{-\theta x_i} = \theta^{an} e^{- \sum_{i=1}^{n} \theta x_i} = \theta^{an} e^{- \theta^n\bar{x}n}$$
since $\sum_{i=1}^{n} x_i = n\bar{x}$
the answer to this is $$\theta^{an} e^{- \theta \bar{x}n}$$
without the $\theta^n$. So where did I go wrong?
$$ ...=\theta^{an} e^{- \sum_{i=1}^{n} \theta x_i} =\theta^{an} e^{- \theta n \frac{1}{n}\sum_{i=1}^{n} x_i} = \theta^{an} e^{- \theta\bar{x}n} $$