Conditional Distribution $X_1\sim {\rm Poi}(\mu_1)\mid \sum X_i$ 
Let $X_i\sim {\rm Poi}(\mu_i)$ independent and $N=\sum_{i=1}^4 X_i$. Show that
  $$X_1\mid N=n\sim {\rm Bin}\left(n,\frac{\mu_1}{\sum\mu_i}\right)$$

I tried without sucess show that
\begin{align}P(X_1|N=n)&=\frac{P(X_1=x_1,N=n)}{P(N=n)}\\[0.2cm]&=\frac{P(X_1=x_1)P(X_2=n-x_1-x_3-x_4)}{P(N=n)}\\[0.2cm]&=\frac{e^{-\mu_1}\mu_1^{x_1}}{x_1!}\frac{e^{-\mu_2}\mu_2^{n-x_1-x_3-x_4}}{(n-x_1-x_3-x_4)!}\frac{n!}{e^{-\sum \mu_i}\sum \mu_1^n}\end{align}
I make some simplifications but I did not get anywhere, maybe I'm taking a wrong path or there is some binomial transformation that I don't see.

That should be:$$\mathsf P(X_1=x\mid N=n) ~=~ \dfrac{\mathsf P(X_1=x)~\mathsf P(X_2+X_3+X_4=n-x)}{\mathsf P(X_1+X_2+X_3+X_4=n)}$$
Now use the fact that the sum of independent Poisson random variables is a Poisson random variable whose parameter is the sum of their parameters.
$$\begin{align}X_2+X_3+X_4 ~\sim~& \mathcal {Poiss}(\mu_2+\mu_3+\mu_4)\\[1ex] X_1+X_2+X_3+X_4~\sim~& \mathcal {Poiss}(\mu_1+\mu_2+\mu_3+\mu_4)\end{align}$$
