In my book, it says:
Independent random variables $X_1, X_2, \dots, X_n$ are modeled by a Poisson distribution with mean $\lambda > 0$. The likelihood for $\lambda$ based on data $\mathbf{x}=(x_1,x_2,\dots\,x_n)^T$ is $$L(\lambda)=\prod_{i=1}^n \frac{\lambda^{x_i}e^{-\lambda}}{x_i!}$$ $$=k \lambda^{n \bar{x}}e^{-n\lambda}$$ where $k$ is a constant.
At first I was a bit confused by this, but I think I get it now. The $e^{-n\lambda}$ bit is easy. For the $\lambda^{n \bar{x}}$ part I think they had $\lambda^{x_1+x_2+\dots,x_n}=\lambda^{\sum{x_i}}$ so $\bar{x}=\frac{\sum{x_i}}{n}$ and $\sum{x_i}=n \bar{x}$ and then $k= \frac{1}{\prod_{i=1}^nx_i!}$ which doesn't depend on $\mathbf{\lambda}$. Have I missed anything ?
