TITLE: Is this equation correct? And if so, is this famous?
QUESTION [5 upvotes]: $$\int_1^{\infty}x^n e^{-x}dx=\frac{1}{e} \sum_{k=0}^n\frac{n!}{(n-k)!}$$

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)

REPLY [8 votes]: 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.