Can anyone explain how this summation was simplified/rearranged? I can't for the life of me follow the steps

Question

See attached image. Part of a question asking to calculate the marginal pmf when the joint pmf is known.

Answer 1

$P(Y=y)=\sum_{z=y}^\infty\binom{z}{y}p^y(1-p)^{z-y}\frac{e^{-\lambda}\lambda ^z}{z!}$.We have $ \binom{z}{y}=\frac{z!}{y!(z-y)!}$ and $\lambda^z=\lambda^y\lambda^{z-y}$ givimg second term.
Let $k=z-y$ so the second term sum becomes $\sum_{k=0}^\infty \frac{(\lambda(1-p))^k}{k!}=e^{\lambda(1-p)}$ giving the third term.

Answer 2

In the first one the summand goes like this$$\binom{z}{y}p^y(1-p)^{z-y}{e^{-\lambda}\lambda^z\over z!}{={z!\over y!(z-y!)}p^y(1-p)^{z-y}{e^{-\lambda}\lambda^z\over z!}\\={1\over y!(z-y!)}p^y(1-p)^{z-y}{e^{-\lambda}\lambda^z\over 1}\\={e^{-\lambda}(p\lambda)^y\over y!}{(1-p)^{z-y}\lambda^{z-y}\over(z-y)!}}$$after all, substitute $u=z-y$ and conclude.