closed-form term for this sum: related to this question:
Is there an easy closed-form term for
$$\sum_{j=k}^{\infty} \frac{x^j}{j!}e^{-x},$$
thus when the sum starts at a constant $k$ instead of $1$?
EDIT:
Thanks for your help. Is there a Chance to solve this sum-term? Because this is not really what I expect, when I talk about a closed-form term. 
A Little bit more of context might help, maybe:
I have $$f(n,p)=\sum_{j=k}^{\infty} \frac{(np)^j}{j!} e^{-np}$$ and it is meant that the partial Derivation is $$\frac{\delta f(n,p)}{\delta n}=\frac{p (np)^{k-1}}{(k-1)!}e^{-np}$$ but I have no idea how to get to this. 
Because to me: 
$$\frac{\delta f(n,p)}{\delta n}=\sum_{j=k}^{\infty} \left( \frac{p (np)^{j-1}}{j!} e^{-np} -\frac{p (np)^j}{j!} e^{-np} \right)$$
but then I am stuck.
 A: $$
\sum_{j=k}^{\infty} \frac{x^j}{j!}e^{-x}=
\left(\sum_{j=k}^{\infty} \frac{x^j}{j!}+\sum_{j=0}^{k-1} \frac{x^j}{j!}-\sum_{j=0}^{k-1} \frac{x^j}{j!}\right)e^{-x}\\
=\left(e^x-\sum_{j=0}^{k-1} \frac{x^j}{j!}\right)e^{-x}=1-e^{-x}\sum_{j=0}^{k-1} \frac{x^j}{j!}
$$
A: Yes:
$$e^{-x}\sum_{j=k}^\infty \frac 1{j!} x^j =
e^{-x}\left[\exp x - \sum_{j=0}^{k-1} \frac 1{j!} x^j \right]
=e^{-x}\int_0^x \frac{(x-t)^{k-1}}{(k-1)!} e^{t} dt
$$
A: You just made a mistake when differentiating $$f(x)=\sum_{j=k}^{\infty} \frac{(xp)^j}{j!} e^{-xp}.$$
The actual derivative is
$$f'(x)=\sum_{j=k}^{\infty} \left( \color{red}{j}\,\frac{p\,(xp)^{j-1}}{j!} e^{-xp} -\frac{p\,(xp)^j}{j!} e^{-xp} \right),$$
that is, provided $k\geqslant1$, using the change of indices $\ell=j-1$ in the first summation,
$$f'(x)=\sum_{\ell=k-1}^{\infty}\frac{p(xp)^{\ell}}{\ell!} e^{-xp} -\sum_{j=k}^{\infty} \frac{p (xp)^j}{j!} e^{-xp}=\frac{p(xp)^{k-1}}{(k-1)!} e^{-xp}.$$
A: Incomplete Gamma function and Eq.2 in : http://mathworld.wolfram.com/IncompleteGammaFunction.html

