Understanding the proof that a probability distribution is Binomial This question is driving me insane as the solution to part (b) makes no sense, I apologize for having to resort to typing out the question and solution - I don't like having to do this, but in this case I make an exception:

You are measuring gamma rays from the decay of a radioactive source.
  The source has a very long half-life so its average rate can be
  assumed to be constant during your measurements. You place a gamma-ray
  detector above the source and take data for a day. You calculate you
  should get an average number $\mu_1$ of gamma rays detected. The
  probability of actually seeing $n_1$ gamma rays in this detector can
  be approximated by a Poisson distribution.
(a) You also place another gamma-ray detector below the source and calculate this should give you an average number $\mu_2$ per day. What
  is the combined probability of seeing $n_1$ in the first detector and
  $n_2$ in the second?
(b) Considering both detectors together as one experiment, then for a given value of the total number of gamma rays observed, $N = n_1
 + n_2$, show that the probability distribution of $n_1$ (or $n_2$) follows a binomial distribution. What is the probability for a single
  trial for this binomial? Does it make sense?


Solution to (a):
Both detectors have a number of hits given by a Poisson distribution, and these are independent. Hence, the combined probability factorises into the product of the two separate probabilities:
$$P(n_1;n_2)=P(n_1;\mu_1)P(n_2;\mu_2)=\cfrac{{\mu_1}^{n_1} e^{-\mu_1}}{n_1!}\cfrac{{\mu_2}^{n_2} e^{-\mu_2}}{n_2!}\tag{1}$$

Solution to (b):
Now consider the total number of observations $N=n_1+n_2$ for a given value of $N$, 
then $n_2=N-n_1$ 
Then from $(1)$ 
$$P(N;n_1)=\cfrac{{\mu_1}^{n_1} e^{-\mu_1}}{n_1!}\cfrac{{\mu_2}^{N-n_1} e^{-\mu_2}}{(N-n_1)!}=\cfrac{{\mu_1}^{n_1} {\mu_2}^{N-n_1}e^{-(\mu_1+\mu_2)}}{n_1!(N-n_1)!}=\cfrac{(\mu_1+\mu_2)^N e^{-(\mu_1+\mu_2)}}{N!}\cfrac{N!}{n_1!(N-n_1)!}\cfrac{{\mu_1}^{n_1} {\mu_2}^{N-n_1}}{(\mu_1+\mu_2)^N}$$
$$\implies P(N;n_1)=\color{red}{\cfrac{(\mu_1+\mu_2)^N e^{-(\mu_1+\mu_2)}}{N!}}\cfrac{N!}{n_1!(N-n_1)!}\left(\cfrac{\mu_1}{\mu_1+\mu_2}\right)^{n_1}\left(\cfrac{\mu_2}{\mu_1+\mu_2}\right)^{N-n_1}\tag{2}$$
$\color{blue}{\fbox{For fixed $N$ the first term (marked red) is constant}}$. The remaining terms are those of a Binomial distribution of $n_1$ outcomes in $N$ trials.
Recalling the Binomial distribution formula: 
$$B(n;p,N)=\cfrac{N!}{n_1!(N-n_1)!}p^n(1-p)^{N-n}\tag{3}$$
where $p$ here is the outcome for a single trial.
Comparing $(2)$ with $(3)$ $\implies p=\cfrac{\mu_1}{\mu_1+\mu_2}$ 
so $1-p=1-\cfrac{\mu_1}{\mu_1+\mu_2}=\cfrac{\mu_1 +\mu_2 -\mu_1}{\mu_1+\mu_2}=\cfrac{\mu_2}{\mu_1+\mu_2}$ as required. The average fraction of the events in the upper detector is $\cfrac{\mu_1}{\mu_1+\mu_2} =
p$, so this makes sense.
End of proof


I understand everything in this proof apart from the part marked $\color{red}{\mathrm{red}}$. I also completely understand what the $\color{blue}{\mathrm{blue} \space \mathrm{box}}$ is saying about the $\color{red}{\mathrm{red}}$ term being constant. But the only way the distribution can be Binomial in order to match $(3)$ is: 
$$\color{red}{\underbrace{\cfrac{(\mu_1+\mu_2)^N e^{-(\mu_1+\mu_2)}}{N!}}_{=1}}\cfrac{N!}{n_1!(N-n_1)!}\left(\cfrac{\mu_1}{\mu_1+\mu_2}\right)^{n_1}\left(\cfrac{\mu_2}{\mu_1+\mu_2}\right)^{N-n_1}$$
But this can't be the case since $P(N;\mu_1+\mu_2)=\color{red}{\cfrac{(\mu_1+\mu_2)^N e^{-(\mu_1+\mu_2)}}{N!}}\ne 1$
Clearly I am missing the point here, so if someone could kindly point it out for me I would be most grateful.  
 A: I am not quite understanding your question, but I will provide my own explanation of this approximation in the hopes that it will help. Let $S_n\sim\operatorname{Bin}(n,p_n)$ and $Y\sim\operatorname{Pois}(\lambda)$ where $\lim_{n\to\infty} np_n = \lambda$. Then the generating functions of $S_n$ and $Y$ are
$$P_n(z) = (1-p_n+p_nz)^n,\quad\quad Q(z) = e^{-\lambda(1-z)}, $$
respectively. In order to prove that $S_n$ converges to $Y$ in distribution, it suffices to show that $P_n$ converges pointwiwse to $Q$ on $(0,1)$. Now, we can write
$$P_n(z) = \left(1 - \frac{np_n(1-z)}n\right)^n=\left(1 - \frac{\lambda_n(1-z)}n\right)^n.$$
where $np_n=\lambda_n$. From L'Hôpital's rule it is clear that
$$\lim_{t\to 0}\frac{\log(1-t)}t=-1. $$
It follows that
$$n\log\left(1 - \frac{\lambda_n}n\right) = \frac{\log(1-\lambda_n/n)}{\lambda_n/n}\lambda_n\stackrel{n\to\infty}\longrightarrow -\lambda.$$
Exponentiating, we have
\begin{align}
\lim_{n\to\infty}P_n(z) &= \lim_{n\to\infty} \left(1 - \frac{\lambda_n(1-z)}n\right)^n\\
&= e^{\lim_{n\to\infty}n\log\left(1-\frac{\lambda_n(1-z)}n \right)}\\
&= e^{-\lambda(1-z)}\\
&= Q(z).
\end{align}
A: Its not enough to do only the substitution $n_1 = N - n_2$. You really need to integrate the second degree of freedom, while minding the proper integration limits and not forgetting the Jacobi determinant.
A: Let $N$  be fixed.  The number $e^{-(\mu_1+\mu_2)}\frac{(\mu_1+\mu_2)^N}{N!}$ cannot be $1$ since it is the probability that a Poisson with parameter $\mu_1+\mu_2$ is equal to $N$, and that is never $1$.  
If $X_1$ and $X_2$ are your two Poisson random variables, then 
$$e^{-(\mu_1+\mu_2)}\frac{(\mu_1+\mu_2)^N}{N!}=\Pr(X_1+X_2=N).$$
To find the conditional distribution of $X_1$, given that $X_1+X_2=N$, divide the expression in your post by the part in red. The conditional distribution of $X_1$ given that $X_1+X_2=N$ is binomial.
However, we can think of $N$ as variable also. Then we get a bivariate distribution which gives us the probability that $X_1+X_2=N$ and $X_1=x_1$. This bivariate distribution is not binomial.
