Is this equation correct? And if so, is this famous? 
$$\int_1^{\infty}x^n e^{-x}dx=\frac{1}{e} \sum_{k=0}^n\frac{n!}{(n-k)!}$$

My proof.1
\begin{align*}
&\int_{1}^{\infty} e^{-\alpha x} \, \mathrm{d}x = \frac{e^{-\alpha}}{\alpha}, \\
&\Rightarrow \qquad \int_{1}^{\infty} (-x)^n e^{-\alpha x} \, \mathrm{d}x = \sum_{k=0}^{n} \frac{n!}{k!(n-k)!} (-1)^k k! \frac{1}{\alpha^k}(-1)^{n-k}e^{-\alpha} \\
&\Rightarrow \qquad \int_{1}^{\infty} x^n e^{-\alpha x} \, \mathrm{d}x = \sum_{k=0}^{n} \frac{n!}{(n-k)!} \frac{e^{-\alpha}}{\alpha^k}.
\end{align*}
$\alpha = 1$とすれば、
\begin{align*}
\int_{1}^{\infty} x^n e^{-x} \, \mathrm{d}x = \frac{1}{e} \sum_{k=0}^{n} \frac{n!}{(n-k)!}.
\end{align*}
が成り立つ。 $\blacksquare$

Original images: (1, 2)

Is it famous?  probalby not.  Is it known?  Yes.
In:
Gradshteyn and Rhyzik, Table of Integrals, Series, and Products
formula 3.351.2 is
$$
\int_u^\infty x^n e^{-\mu x} dx = e^{-u \mu} \sum_{k=0}^n \frac{n!}{k!}\;
\frac{u^k}{\mu^{n-k+1}}
$$
Now if you take $u=\mu=1$ you get your formula.
