Simplification. Difficulty in detecting the series $\sum_{i=0}^\infty \frac{x^i}{i!} = e^x$

Question

Any suggestion on how to simplify the following?

$$\sum_{m=k+1}^\infty \frac{((1-p)\lambda)^m}{(m-k)!}$$

Answer 1

Hint: Let $m-k=z\Rightarrow m=z+k$. Then

$$\sum_{m=k+1}^\infty \frac{((1-p)\lambda)^m}{(m-k)!}=\sum_{z=1}^{\infty} \frac{((1-p)\cdot \lambda)^{z+k}}{z!}=((1-p)\cdot \lambda)^{k}\cdot \sum_{z=1}^{\infty} \frac{((1-p)\cdot \lambda)^{z}}{z!}$$

Answer 2

Hint: Write out the first few terms:

\begin{align} \sum_{m=k+1}^\infty \frac{((1-p)\lambda)^m}{(m-k)!} &= \frac{((1-p)\lambda)^{k+1}}{1!}+\frac{((1-p)\lambda)^{k+2}}{2!}+\frac{((1-p)\lambda)^{k+3}}{3!}+\dots\\ &= ((1-p)\lambda)^k\left(\frac{((1-p)\lambda)^{1}}{1!}+\frac{((1-p)\lambda)^{2}}{2!}+\frac{((1-p)\lambda)^{3}}{3!}+\dots\right) \\ &= ((1-p)\lambda)^k\left(-1+1+\frac{((1-p)\lambda)^{1}}{1!}+\frac{((1-p)\lambda)^{2}}{2!}+\frac{((1-p)\lambda)^{3}}{3!}+\dots\right) \\ &= ((1-p)\lambda)^k\left(-1+e^{(1-p) \lambda}\right) \\ \end{align}