# likelihood of poisson distribution

If $$(x_1,\cdots, x_n)$$ is a sample from a Poisson(θ) distribution, where $$θ ∈ (0,∞)$$ is unknown, then determine the $$MLE$$ of $$θ$$.

my attempt:

so the probability density of poisson is

$$p_{\theta}(x) = \frac{\theta^x e^{-\theta}}{x!}$$

the likelihood is given by

$$L(\theta | \bar{x}) = \prod_{i=1}^{n} \frac{\theta^{x_i}e^{-\theta}}{x_i!} = e^{-n\theta} \left(\prod_{i=1}^{n} \frac{1}{x_i!} \right) \theta^{\sum_{i=1}^{n}x_i}$$

How do I do the factorial ?

For some reason in the solution their answer is just $$e^{-n\theta} \theta^{\sum_{i=1}^{n}x_i}$$ ?

The factorial part is a (positive) constant (with respect to the parameter $$\theta$$), so when maximising the likelihood with respect to $$\theta$$, it can be ignored. (It is still technically part of the likelihood though.)
It's like if you wanted to find $$\theta$$ to maximise a function $$5f(\theta)$$, it's the same as just finding the maximiser of $$f(\theta)$$ (since $$5$$ is a positive constant).
(Remember, when we find maximum likelihood estimators, we maximise the likelihood or log-likelihood with respect to the parameter $$\theta$$, and treat the $$x_i$$ as constants.)