From my textbook after discussing the Gamma distribution and using the Poisson process postulates:
It is left as an exercise to verify that
$$ \int_{\lambda w}^{\infty}\frac{z^{k-1}e^{-z}}{(k-1)!} \ dz=\sum_{x=0}^{k-1}\frac{(\lambda w)^xe^{-\lambda w}}{x!} $$
Here's what I have so far:
\begin{align} \int_{\lambda w}^{\infty}\frac{z^{k-1}e^{-z}}{(k-1)!} \ dz&= \int_{0}^{\infty}\frac{z^{k-1}e^{-z}}{(k-1)!} \ dz - \int_{0}^{\lambda w}\frac{z^{k-1}e^{-z}}{(k-1)!} \ dz \\ &= \frac{1}{(k-1)!} \left[\Gamma(k) - \int_0^{\lambda w}z^{k-1}e^{-z} \ dz \right] \end{align}
I don't know if I set this up in my favor, but that right hand integral is getting nowhere with integration by parts. Am I missing something clever that I need to do?
For ease of notation let $\mu = \lambda w$.
Note that integration by parts yields $$\int_\mu^\infty z^{k-1} e^{-z} \, dz = \mu^{k-1} e^{-\mu} + (k-1) \int_\mu^\infty z^{k-2} e^{-z} \, dz.$$ By a nearly identical argument, the integral on the right can be shown to be $$\int_\mu^\infty z^{k-2} e^{-z} \, dz = \mu^{k-2} e^{-\mu} + (k-2) \int_\mu^\infty z^{k-3} e^{-z} \, dz,$$ and you can continue integrating by parts on, until you end up with the base case $\int_\mu^\infty e^{-z} \, dz = e^{-\mu}$.
Thus, $$\int_\mu^\infty z^{k-1} e^{-z} \, dz = e^{-\mu} (\mu^{k-1} + (k-1) \mu^{k-2} + (k-1)(k-2) \mu^{k-3} + \cdots + (k-1)!) = e^{-\mu} \sum_{m=0}^{k-1} \mu^m \frac{(k-1)!}{m!}.$$
Dividing both sides by $(k-1)!$ leads to the desired equality.
Approach using Poisson process with rate $\lambda$:
The waiting time until the $k$th arrival follows the gamma distribution whose PDF is the integrand on the left-hand side, so the left-hand side is the probability that the $k$th arrival occurs after time $w$.
The number of arrivals in the time interval $[0, w]$ follows the Poisson distribution with rate $\lambda w$, so the right-hand side is the probability that $< k$ arrivals occur in the time interval $[0, w]$.
These are the same.