I am stuck with an arithmetic detail of a proof of Sum of Independent Poissons.
\begin{align*} P(X+ Y =k) &= \sum_{i = 0}^k P(X+ Y = k, X = i)\\ &= \sum_{i=0}^k P(Y = k-i , X =i)\\ &= \sum_{i=0}^k P(Y = k-i)P(X=i)\\ &= \sum_{i=0}^k e^{-\mu}\frac{\mu^{k-i}}{(k-i)!}e^{-\lambda}\frac{\lambda^i}{i!}\\ &= e^{-(\mu + \lambda)}\frac 1{k!}\sum_{i=0}^k \frac{k!}{i!(k-i)!}\mu^{k-i}\lambda^i\\ \end{align*}
The thing I don't get in the last equation is where does this ${k!}$ in the numerator come from? I would assume that at this point it should rather be:
\begin{align*} e^{-(\mu + \lambda)}\frac 1{k!}\sum_{i=0}^k \frac{1}{i!(k-i)!}\mu^{k-i}\lambda^i\\ \end{align*}
But it has to be
\begin{align*} e^{-(\mu + \lambda)}\frac 1{k!}\sum_{i=0}^k \frac{k!}{i!(k-i)!}\mu^{k-i}\lambda^i\\ \end{align*}
Since the next step in the proof is:
\begin{align*} e^{-(\mu + \lambda)}\frac 1{k!}\sum_{i=0}^k \binom ki\mu^{k-i}\lambda^i\\ \end{align*}
Followed by:
\begin{align*} \frac{(\mu + \lambda)^k}{k!} \cdot e^{-(\mu + \lambda)} \end{align*}
By the way, the last step is also a bit cryptic to me. I don't really see how $\sum_{i=0}^k \binom ki\mu^{k-i}\lambda^i$ becomes $(\mu + \lambda)^k$.
EDIT: Nevermind the second doubt of mine. It's the Binomial theorem, clearly.
I tried to consult a number of other sources, but the proofs don't go to this level of detail, thus I am asking for your help. The above notation is borrowed directly from this post, which already is quite detailed.
