Integration is (fundamentally) summation

Nice little problem!

$\displaystyle \int _0^1 {x^{2015} e^{-x} \ dx} =2015! - k \sum_{r=0}^{2015}{\dbinom{2015}{r} r!}$

$\displaystyle \int _0^1 {x^{2015} e^{-x} \ dx} + k \sum_{r=0}^{2015}{\dbinom{2015}{r} \Gamma(r+1)} = 2015!$

$\displaystyle \int _0^1 {x^{2015} e^{-x} \ dx} + k \sum_{r=0}^{2015}{\dbinom{2015}{r} \int_0^{\infty}{x^r e^{-x} \, dx}} = 2015!$

$\displaystyle \int _0^1 {x^{2015} e^{-x} \ dx} + k \int_0^{\infty}{\sum_{r=0}^{2015}{\dbinom{2015}{r} x^r} e^{-x} \, dx} = 2015!$

$\displaystyle \int _0^1 {x^{2015} e^{-x} \ dx} + k \int_0^{\infty}{(1+x)^{2015} e^{-x} \, dx} = 2015!$

$\displaystyle \int _0^1 {x^{2015} e^{-x} \ dx} + k \int_1^{\infty}{x^{2015} e^{-(x-1)} \, dx} = 2015!$

$\displaystyle \int _0^1 {x^{2015} e^{-x} \ dx} + k \times e \int_1^{\infty}{x^{2015} e^{-x} \, dx} = 2015!$

We know however that $\displaystyle 2015! = \int_0^{\infty}{x^{2015} e^{-x} \, dx} = \int_0^1 {x^{2015} e^{-x} \, dx} + \int_1^{\infty}{x^{2015} e^{-x} \, dx}$

Substituting it, we get $\displaystyle k \times e = 1 \Rightarrow \large k = \frac{1}{e}$

We can continuously indefinitely integrate the l.h.s of the question. writing every step of this by parts one below one we can cancel the lhs. Then simplifying and putting limit we get a series. The series is nothing but the r.h.s. I am uploading a pic.sorry for handwriting .

Much more easy without gamma function and proving

$I(n)=-\frac { 1 }{ e } +nI(n-1)\\ where\quad I(n)=\int _{ 0 }^{ 1 }{ { x }^{ n } } { e }^{ -x }dx$