How can a conditional law of a random value following a poisson law be a Binomial law?

Let say that $X_1,\dots ,X_m$ are independent random variables following Poisson law of parameter $λ_1,\dots, λ_m$.

I'm looking for the the conditional law of $X_1$ conditionated by $\{X_1+X_2=k\}$ which is said to be a Binomial law $B(k,\frac{\lambda_1}{\lambda_1+\lambda_2})$.

Yet, $\sum\limits_{k\in X_2(\Omega)}{P(X_2=k-j)}$ has sense while $k-j\ge 0$.

Thus \begin{align*} \sum_{k\in X_2(\Omega)}{P(X_2=k-j)}&=\sum\limits_{k-j>=0}{\frac{\lambda_2^{k-j}e^{-\lambda_2}}{(k-j)!}}\\[10pt] &= e^{-2\lambda_2} \end{align*}

Therefore, we can calculate $(PX_1=j\mid X_1+X_2=k)$

\begin{align*} (PX_1=j\mid X_1+X_2=k)&=\frac{\frac{\lambda_1e^{-\lambda_1-2\lambda_2}}{j!}}{\frac{(\lambda_1+\lambda_2)^ke^{-\lambda_1+\lambda_2}}{k!}} \end{align*}

Here I'm stuck, I don't think I can find a Binomial law from here... Have you any hint? Did I di a mistake?

Hint: For the top, you want $\Pr(X_1=j \cap X_2=k-j)$. No summation, $k$ is fixed.
This is $$e^{-\lambda_1}\frac{\lambda_1^j}{j!}e^{-\lambda_2}\frac{\lambda_2^{k-j}}{(k-j)!}.$$ Now there will be nice cancellation. If trouble persists, I can finish the calculation.