I am trying to construct a likelihood function for a truncated poisson distribution, the formula for my probability mass function is: $$f_X(x)=\frac{\lambda^x}{(e^{\lambda}-1)(x!)}$$
I know that in order to construct a likelihood function I need to take the product of each instance of $x_i$ from 1 to n, i.e. $$\prod_{i=1}^{n} \frac{\lambda^x}{(e^{\lambda}-1)(x!)}$$
My only problem is that I can't work out how to handle the $x!$ term in my working. Thus far, I have:
$$ L(\lambda)=\frac{\lambda^{\sum(x_{i})}}{(e^\lambda-1)^n(\prod{x!})}$$
This doesn't feel complete though as I feel the product over factorials term can be somehow simplified using the gamma function. I'm going to need to take the log of my likelihood function later and differentiate it and it's a bit awkward with the factorials stuff.
Would really appreciate any help! I'm very new to stats and a push in the right direction would be lovely.
$\prod{x!}$ does not contain $\lambda$ and so is not important in considering the shape of the likelihood function
From your expression, the logarithm of the likelihood is
$$\log(L(\lambda))=\left(\sum x_{i}\right)\log(\lambda)-n\log(e^\lambda-1) -\log\left(\prod{x!}\right)$$
and its derivative with respect to $\lambda$ is
$$\frac{d\log(L(\lambda))}{d\lambda}=\frac{\sum x_{i}}{\lambda}-\frac{n}{1-e^{-\lambda}}$$
at which point the $\prod{x!}$ term has disappeared
If instead you had taken the derivative of the likelihood function, it would still be there but only as a constant of proportionality but, for example, would be irrelevant if finding the maximum likelihood and would be subsumed in to the normalising calculation if finding a Bayesian posterior distribution.