Independent random variables distributed by Poisson law The problem: 
Let $\xi$ and $\eta$ be independent random variables distributed by Poisson law with parameters $\lambda_1$ and $\lambda_2$ correspondingly. Show that random variable $\xi + \eta$ also distributed by Poisson (with parameters  $\lambda{1} +  \lambda{2}$)
My attempt:
By Poisson law  $P(\xi + \eta ; \lambda_1 +  \lambda_2) = \frac{(\lambda_1 +  \lambda_2)^{\xi + \eta} * e^{-(\lambda_1 +  \lambda_2)}}{(\xi + \eta)!}$
. And since we know that they are independent we can say that $P(\xi + \eta ; \lambda_1 +  \lambda_2)$ is also equal to $P(\xi ; \lambda_1) + P(\eta ; \lambda_2) - P(\xi ; \lambda_1) * P(\eta ; \lambda_2)$. 
I tried simplifying the long equation that I get but I still cannot get them equal to each other. Should I try simplifying them from both sides? 
 A: For greater clarity, let's use Latin capital letters for random variables, lower case for their values, and Greek letters for the parameters so the question becomes

Let $X$ and $Y$ be independent random variables distributed by Poisson law with parameters $\lambda_1$ and $\lambda_2$ correspondingly. Show that random variable $Z=X+Y$ is also distributed by Poisson (with parameters  $\lambda_{1} +  \lambda_{2}$)

Your "$P(\xi ; \lambda_1) + P(\eta ; \lambda_2) - P(\xi ; \lambda_1) * P(\eta ; \lambda_2)$" is related to independent events rather than to random variables.  
A better approach here would be to look at something like $$\mathbb{P}(Z=z)=\sum_x \mathbb{P}(X=x)\, \mathbb{P}(Y=z-x)$$ which since $x$ and $z-x$ must be non-negative integers gives $$\mathbb{P}(Z=z)=\sum_{x=0}^{z} \lambda_1^x \frac{e^{-\lambda_1}}{x!}\, \lambda_2^{z-x} \frac{e^{-\lambda_2}}{(z-x)!}$$
You want this to be equal to $(\lambda_1+\lambda_2)^z \, \dfrac{e^{-(\lambda_1+\lambda_2)}}{z!}$. I will leave that to you (hint: think binomial expansion).
