Proof of the expression of upper incomplete gamma function as an finite summation. I am trying the proof the one of the finite summation expression of the incomplete upper Gamma function
$$ \Gamma(m,x) = \int_{x}^\infty v^{m-1}e^{-v}dv$$
So, using the equality (also given in Tab. of. Int. [3.351.2])
$$\int_{u}^{\infty} x^{n}e^{-\mu x}dx = e^{-\mu u} \sum_{k=0}^{n} \frac {n!}{k!} \frac{u^k}{\mu^{n-k+1}} $$
I can also express the gamma function in terms of a finite summation.
So, I'll be so glad if you help me to prove the equality given just above.
 A: $$\int_{u}^{+\infty}x^{n}e^{-\mu x}dx\stackrel{x=t+u}{=}e^{-\mu u}\int_{0}^{+\infty}\left(t+u\right)^{n}e^{-\mu t}dt$$ $$\stackrel{\mathrm{Bin}}{=}e^{-\mu u}\sum_{k=0}^{n}\dbinom{n}{k}u^{k}\int_{0}^{+\infty}t^{n-k}e^{-\mu t}dt\stackrel{t=v/\mu}{=}e^{-\mu u}\sum_{k=0}^{n}\dbinom{n}{k}\frac{u^{k}}{\mu^{n-k+1}}\int_{0}^{+\infty}v^{n-k}e^{-v}dt$$ $$=\color{red}{e^{-\mu u}\sum_{k=0}^{n}\frac{n!}{k!}\frac{u^{k}}{\mu^{n-k+1}}}$$
A: Here is a sketch of a simple approach.  Write
$$ f(u) = e^{\mu u} \int_u^\infty x^n e^{-\mu x} dx $$
Differentiating with respect to $u$ gives, by the product rule and the fundamental theorem of calculus,
$$ f'(u) = \mu e^{\mu u} \int_u^\infty x^n e^{-\mu x} dx - e^{\mu u} u^n e^{-\mu u} = \mu f(u) - u^n $$
This can be used as a recursion relation:
$$ f^{(2)}(u) = \mu^2 f(u) - \mu n u^{n-1} - u^n $$
$$ f^{(3)}(u) = ... $$
If you take the Taylor series
$$ \sum_k \frac{f^{(k)}(0) u^k}{k!} $$
and plug in the fact that $f(0)=\frac{n!}{\mu^{n+1}}$ then you basically get the desired result above.  When $k>n$, then there will always be one of the terms that originally was $u^n$ but has received $n$ derivatives, so it will not vanish when you plug in $u=0$, and it will cancel with the term containing $f(0)$, which is why the series terminates.
