Bayesian: Poisson with constant prior Need some help.
Assume the sampling distribution is Poisson with sample size $n$.
Assume constant prior distribution.
Likelihood:
$L(\lambda | \mathbb{x})=\prod_{i=1}^{n}\dfrac{e^{-\lambda}\lambda^{x_{i}}}{x_{i}!}=\dfrac{e^{-n\lambda}\lambda^{\sum_{i=1}^{n}x_{i}}}{\prod_{i=1}^{n}x_{i}!}$
Prior:
$p(\lambda)=\frac{1}{a}$, $a$ is some real constant.
Therefore, posterior:
$\pi(\lambda | \mathbb{x})=\dfrac{e^{-n\lambda}\lambda^{\sum_{i=1}^{n}x_{i}}}{\prod_{i=1}^{n}x_{i}!}\cdot\dfrac{1}{a}$
$\pi(\lambda | \mathbb{x})\propto e^{-n\lambda}\lambda^{\sum_{i=1}^{n}x_{i}}$
Is this correct? And if so, is there no nice posterior for the Poisson/constant?
 A: This looks mostly reasonable to me, but your terminology is a bit loose, and I believe there is also one mathematical oversight. My understanding of the problem would be as follows.
First, applying Bayes theorem to the Poisson parameter $\lambda$ and the data $\mathbb{x}$ gives the posterior formally as
$$p(\lambda \mid \mathbb{x})=\frac{p(\mathbb{x} \mid \lambda)p(\lambda)}{p(\mathbb{x})} \,,\, p(\mathbb{x})=\int_0^{\infty}{p(\mathbb{x} \mid \lambda)p(\lambda)d\lambda}$$
where the integral for the normalization factor $p(\mathbb{x})$ is over all possible $\lambda\in\mathbb{R}^+$.
Then, assuming i.i.d. data $\mathbb{x}\in\mathbb{N}^n$, your expression for the data likelihood appears to be correct, and is equivalent to
$$p(\mathbb{x} \mid \lambda)=\frac{(e^{-\lambda}\lambda^{\bar{x}})^n}{\Omega}\,,\, \bar{x}=\frac{1}{n}\sum_{i=1}^nx_i\,,\,\Omega=\prod_{i=1}^{n}x_{i}!$$
where $\Omega$ just collects the factors that do not depend on $\lambda$.
Next we must consider the prior $p(\lambda)$, and this is where I believe your mathematical oversight occurs. For a generic Poisson distribution, we know $\lambda\in\mathbb{R}^+$. The prior you give, when interpreted literally, does not give a valid PDF over $\mathbb{R}^+$ (i.e. it integrates to $\infty$). So really the implied prior is a uniform PDF over $[0,a]$, i.e.
$$p(\lambda)=\frac{\mathbb{1}_{\lambda\in[0,a]}}{a}$$
where the notation $\mathbb{1}_C$ is an indicator function that is $1$ when condition $C$ holds, and $0$ otherwise.
Combining the above, the posterior would then be
$$p(\lambda \mid \mathbb{x})=\frac{(e^{-\lambda}\lambda^{\bar{x}})^n}{\int_0^a{(e^{-\lambda}\lambda^{\bar{x}})^nd\lambda}}\mathbb{1}_{\lambda\in[0,a]}$$
(Warning: I am probably out of my depth, so may be wrong here!) 
A: The posterior with such an improper prior is a gamma distribution and you can pretty much read off the parameter values from what you wrote down. And yes, your derivation seems right.
A: The likelihood function can be write as:
$L(\mathbb{x}|\lambda) \propto \prod_{i=1}^N e^{\lambda}x^{x_i}$
$\propto$ means you can ignore the normalize constant at this step. 
Prior: $p(\lambda) \propto \frac{1}{a} \propto 1$
Therefore, the posterior distribution is:
$\pi(\lambda | \mathbb{x}) \propto L(\mathbb{x}|\lambda) \times p(\lambda)$
$\pi(\lambda | \mathbb{x}) \propto e^{N\lambda}x^{\sum_{i=1}^Nx_i}$
If you are going to just write down the Bayes rule, then it is like this. But if you want to implement MCMC, then you need to consider if a non-informative prior is proper.
