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)
 A: It depends on what it means to be famous, but it is a quite well-known formula in probability theory. Indeed, let $X$ be the sum of $(n+1)$ i.i.d. $\text{Exp}(1)$ random variables. Then

*

*$X$ has the PDF $f_X(x) = (x^n e^{-x} / n!) \mathbf{1}(x > 0)$, and so,
$$ \int_{1}^{\infty} x^n e^{-x} \, \mathrm{d}x = n! \,\mathbf{P}(X > 1). $$


*$X$ can be realized as the $(n+1)$-th arrival time of the Poisson process $N = (N_t)_{t\geq 0}$ with unit rate, and so,
\begin{align*}
\mathbf{P}(X > 1)
&= \mathbf{P}(\text{$(n+1)$-th arrival has not occurred by time $1$}) \\
&= \mathbf{P}(\text{there are at most $n$ arrivals by time $1$}) \\
&= \mathbf{P}(N(1) \leq n) = \sum_{k=0}^{n} \frac{1}{k!}e^{-1}. \end{align*}
Combining two observations proves the desired identity.
A: It is true and changing the summation order results in a slightly different expression, which I find easier to prove:

$\int_{1}^{\infty} x^n e^{-x}dx = \frac{1}{e}\sum_{k = 0}^{n} \frac{n!}{k!}$

You can prove it by using induction and integration by parts:

case n = 0:
$\int_{1}^{\infty} x^0 e^{-x}dx = [-e^{-x}]_1^\infty = 0 + \frac{1}{e} = \frac{1}{e} \frac{0!}{0!}$
step $n \rightarrow n+1$:
$\int_{1}^{\infty} x^{n+1} e^{-x}dx = [-x^{n+1}e^{-x}]_1^\infty + (n+1)\int_{1}^{\infty} x^{n} e^{-x}dx = \frac{1}{e} + (n+1)\left(\frac{1}{e}\sum_{k = 0}^{n} \frac{n!}{k!} \right) = \frac{1}{e} \left(1 + \sum_{k = 0}^{n} \frac{(n+1)!}{k!}\right) = \frac{1}{e}\sum_{k = 0}^{n+1} \frac{(n+1)!}{k!}$

A: 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.
