Reference-request: Closed form of $\int_0^1 (\operatorname{W_{-1}}(-\tfrac t{\mathrm{e}}))^n \, dt$ $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\floor{\operatorname{floor}}$
\begin{align}
\int_0^1 (\Wm(-\tfrac t{\e}))^n \, dt
&=(-1)^n\sum_{k=0}^n (k+1)!\binom{n}{k}
\tag{1}\label{1}
,
\end{align} 
where $\Wm$ is the real branch of the Lambert $\W$ function.
A001339 also suggests that
\begin{align}
\int_0^1 (\Wm(-\tfrac t{\e}))^n \, dt
&=(-1)^n
(\floor(\e\cdot n\cdot n!) + 1)
\tag{2}\label{2}
.
\end{align} 
Question: Are \eqref{1},\eqref{2} known relations? Any reference?
$\endgroup$
 A: I cannot give you any reference requested for this interesting result you give, but I can at least give a reason as to why your definite integral is equal to the term involving the floor function $\lfloor x \rfloor$.
From your result
$$I_n = \int_0^1 \left [\operatorname{W}_{-1} \left (-\frac{t}{e} \right ) \right ]^n \, dt = (-1)^n \sum_{k = 0}^n (k + 1)! \binom{n}{k}, \quad n = 0,1,2,\ldots \tag1$$
I will first show that
$$I_n = \frac{(-1)^n}{n + 1} \big{(}e n \Gamma (n + 2,1) + 1 \big{)}.$$
Here $\Gamma (s,x)$ is the upper incomplete gamma function.
From the exponential sum function, namely
$$\exp_n (x) = \sum_{k = 0}^n \frac{x^k}{k!} =  \frac{e^x \Gamma (n + 1, x)}{\Gamma (n + 1)},$$
setting $x = 1$ and shifting the index $n$ to $n + 1$ gives
$$\sum_{k = 0}^{n + 1} \frac{1}{k!} = \frac{e \Gamma (n + 2,1)}{\Gamma (n + 2)} = \frac{e \Gamma (n + 2,1)}{(n + 1)n!}.$$
Rearranging, this can be written as
$$\frac{e n \Gamma (n + 2,1) + 1}{n + 1} = \sum_{k = 0}^n \frac{n n!}{k!} + 1 \tag2$$
Returning to the finite sum in (1), reindexing $k \mapsto n - k$ we have
$$\sum_{k = 0}^n (k + 1)! \binom{n}{k} = \sum_{k = 0}^n (n - k + 1)! \binom{n}{n - k}.$$
And as $\binom{n}{k} = \binom{n}{n - k}$ we have
\begin{align}
\sum_{k = 0}^n (k + 1)! \binom{n}{k} &= \sum_{k = 0}^n (n - k + 1) (n - k)! \binom{n}{k}\\
&= \sum_{k = 0}^n (n - k + 1) \frac{n!}{k!}\\
&= \sum_{k = 0}^n \frac{n n!}{k!} - \sum_{k = 0}^n \frac{(k - 1) n!}{k!}\\
&= \sum_{k = 0}^n \frac{n n!}{k!} - S_1.
\end{align}
For the sum $S_1$
\begin{align}
S_1 &= \sum_{k = 0}^n \frac{(k - 1) n!}{k!}\\
&= -n! + n! \sum_{k = 2}^n \frac{k - 1}{k!}\\
&= -n! + n! \sum_{k =2}^n \left (\frac{1}{(k - 1)!} - \frac{1}{k!} \right )\\
&= -n! + n! \left (1 - \frac{1}{n!} \right )\\
&= -1.
\end{align}
So we can immediately see that
$$\sum_{k = 0}^n (k + 1)! \binom{n}{k} = \sum_{k = 0}^n \frac{n n!}{k!} + 1 = \frac{e n \Gamma (n + 2,1) + 1}{n + 1}.$$
Thus
$$I_n = \frac{(-1)^n}{n + 1} \big{(}e n \Gamma (n + 2,1) + 1 \big{)}. \tag3$$
Next, noting that (see here)
$$\Gamma (s + 1,1) = \frac{\lfloor e s! \rfloor}{e},$$
where $s$ is a positive integer, one can rewrite (3) as
$$I_n = \frac{(-1)^n}{n + 1} \big{(}n \lfloor e(n + 1)! \rfloor + 1 \big{)}.$$
