Derivative of a (complicated) sum Given
$$
f(x)=e^{-ax}\ \sum_{k=0}^{r-1}\frac{(ax)^k}{k!}
$$
How do I show that
$$
f'(x)= -\frac{a^r}{(r-1)!}x^{r-1}e^{-ax}
$$
Thank you in advance!

It is:
$$f(x)=e^{-ax}\ \sum_{k=0}^{r-1}\frac{(ax)^k}{k!}=e^{-ax}\left(e^{ax}-\sum_{k=r}^{\infty} \frac{(ax)^k}{k!}\right)=1-e^{-ax}\sum_{k=r}^{\infty}\frac{(ax)^k}{k!} \Rightarrow \\
f'(x)=\color{red}{ae^{-ax}}\sum_{k=r}^{\infty}\frac{(ax)^k}{k!}-\color{red}{ae^{-ax}}\sum_{k=r}^{\infty}\frac{(ax)^{k-1}}{(k-1)!}=\color{red}{ae^{-ax}}\cdot \left(-\frac{(ax)^{r-1}}{(r-1)!}\right)=-\frac{a^r}{(r-1)!}x^{r-1}e^{-ax}.$$
