Proof using binomial coeff I don't understand the step between left side and right side of my ?
I 
 A: Bring the $e^{-a}e^{-b}$ "out" and divide and multiply by $k!$. We get
$$\frac{e^{-a}e^{-b}}{k!} \sum_{j=0}^k \frac{k!}{j!(k-j)!}a^jb^{k-j}.$$
This can be rewritten as 
$$\frac{e^{-a}e^{-b}}{k!} \sum_{j=0}^k \binom{k}{j}a^jb^{k-j}.$$
We recognize that the inner sum is the binomial expansion of $(a+b)^k$. So our expression is equal to
$$\frac{e^{-a}e^{-b}}{k!}(a+b)^k.$$
The derivation quoted is a little more complicated. The author multiplied and divided by $(a+b)^k$, and got an inner sum of 
$$\sum{j=0}^k \binom{k}{j}\left(\frac{a}{a+b}\right)^j \left(\frac{b}{a+b}\right)^{k-j}.$$
The inner sum is then the binomial expansion of $\left(\frac{a}{a+b}+\frac{b}{a+b}\right)^k$, that is, of $1^k$, so it is $1$. 
A: A slightly different presentation:
$$
\begin{align}
\frac{a^je^{-a}}{j!}\frac{b^{k-j}e^{-b}}{(k-j)!}
&=\frac{\color{#C00000}{(a+b)^k}e^{-(a+b)}}{\color{#00A000}{k!}}\frac{\color{#00A000}{k!}}{j!\,(k-j)!}\frac{a^j}{\color{#C00000}{(a+b)^j}}\frac{b^{k-j}}{\color{#C00000}{(a+b)^{k-j}}}\\
&=\frac{(a+b)^ke^{-(a+b)}}{k!}\binom{k}{j}\left(\frac{a}{a+b}\right)^j\left(\frac{b}{a+b}\right)^{k-j}
\end{align}
$$
Of course, we also have
$$
\sum_{j=0}^k\binom{k}{j}\left(\frac{a}{a+b}\right)^j\left(\frac{b}{a+b}\right)^{k-j}
=\left(\frac{a}{a+b}+\frac{b}{a+b}\right)^k=1
$$
