statistics - Sum in Probability $sum_{n=k}^infty frac{(lambdacdot s)^n}{n!}e^{-lambda s} binom{n}{n-k}p^{n-k}(1-p)^k$

I am trying to evaluate a sum consisting of a possion distribution, multiplied by a binomial distribution. I already know by the power of maple that

$$\sum_{n=k}^\infty \frac{(\lambda\cdot s)^n}{n!}e^{-\lambda s} \binom{n}{n-k}p^{n-k}(1-p)^k = \frac{(s\lambda)^ke^{-s\lambda}}{k!} (1-p)^k e^{s\lambda p}$$

Where the answer also could be written as $P_r(s\lambda)(1-p)^k e^{s\lambda p}$.
My guess is that I have to split the sum into $[k,\infty)=[0,\infty)-[0,k-1]$.

Also the sum reminds me of one of the definitions of $e$, and the binomial formula
$$e^x = \sum_{k}\frac{x^k}{k!} \ , \qquad (x+y)^n=\sum_k \binom{n}{k}x^{n-k}y^k$$
Where both sums goes from $0$ to $\infty$. However I am not quite able to see how to use this to evalutate the sum. Any help or advice is greatly appreaciated.

For the curious I am trying to solve 6b from here, it should be
$$P(Y=k) = \sum_{n=k}^\infty P(Y=k \cap N=n) = \frac{(s\lambda)^ke^{-s\lambda}}{k!} (1-p)^k e^{s\lambda p}$$

Answer

Factoring the term $\displaystyle (\lambda s)^k\mathrm e^{-\lambda s}\frac1{k!}(1-p)^k$ which is independent of $n$, one is left with the series

$$(\lambda s)^k\mathrm e^{-\lambda s}\frac1{k!}(1-p)^k\cdot\sum_{n\geqslant k}(\lambda s)^{n-k}\frac1{(n-k)!}p^{n-k}.$$
You might want to finish, identifying the last sum...