How-to proof this integral I saw it in the Hurwitz zeta function ,
$$ \int_0^ty^{p-1}\left(1-e^{-zy}\right)dy=\frac{t^p}{p}+e^{-tz}\sum_{k=0}^{p-1}k!\binom{p-1}{k}\frac{t^{p-1-k}}{z^{k+1}}-\frac{(p-1)!}{z^p}$$
And I was not sure of the second term of right hand side.Any help is appreciated
 A: Hint. One may start with
$$
\int_0^te^{\large -zy}\:dy=\frac{1-e^{\large -zt}}z, z>0
$$ then differentiating $p$ times with respect to $z$, using the Leibniz formula, one gets
$$
\begin{align}
\int_0^t(-y)^{p}e^{-zy}\:dy&=\frac{\partial^{p}}{\partial z^{p}}\left(\frac{1-e^{\large -zt}}z\right)
\\\\&=\sum_{k=0}^{p}{\binom{p}k}\frac{\partial^{k}}{\partial z^{k}}\left(\frac1z\right)\cdot\frac{\partial^{p-k}}{\partial z^{p-k}}\left(1-e^{\large -zt}\right)
\\\\&=\sum_{k=0}^{p}{\binom{p}k}\frac{(-1)^kk!}{z^{k+1}}\cdot  (-1)^{k}t^{p-k} \cdot e^{-zt}
\end{align}
$$ as announced.
A: For any $\rho > 0$, it is true that
$$
\begin{aligned}
\int_0^t -y^{\rho - 1}e^{-zy}\mathrm{d}y &= \int_0^t \frac{y^{\rho-1}}{z}(-ze^{-zy})\mathrm{d}y
\\
&= \left(\frac{e^{-zy}}{z}y^{\rho-1}\right)_{y=0}^t + \frac{(\rho-1)}{z}\int_0^t -y^{\rho-2}e^{-yz}\mathrm{d}y
\\
&= \frac{e^{-tz}}{z}t^{\rho-1} + \frac{(\rho-1)}{z}\int_0^t -y^{\rho-2}e^{-yz}\mathrm{d}y.
\end{aligned}
$$
Collecting all the terms, you obtain the described series.
$$
\frac{e^{-tz}}{z}t^{\rho-1} + (\rho-1)\frac{e^{-tz}}{z^2}t^{\rho-2} + \dotsb + \prod_{k=1}^{\rho-1}(\rho-k)\frac{e^{-tz}}{z^{\rho}} = e^{-tz}\left(\sum_{k=0}^{\rho - 1}\binom{\rho-1}{k}\frac{t^{\rho-1-k}}{z^{k+1}} - \frac{(\rho-1)!}{z^\rho}\right).
$$
