In my answer to this question, $W(.)$ being Lambert function, I indirectly showed that $$W_0(x)-W_{-1}(x)=1 \implies x=-\frac {1} {e-1}\, \exp \left( \frac {-1} {e-1}\right)$$ Is there any way to prove it directly ?
Edit after Szeto's answer
Back here, this means that $$W_0(x)=\frac{1}{1-e} \qquad \text{and} \qquad W_{-1}(x)=\frac{e}{1-e}$$ which do not seem do appear as special values of Lambert function.
Using the equation (1)(second page of the paper)
If $d=W_0(z)-W_{-1}(z)$, $$\frac{d}{e^{-d}-1}\text{exp}\left(\frac{d}{e^{-d}-1}\right)=z$$
Here, $d=1$, and your result follows:
$$z=-\frac{e}{e-1}\text{exp}\left(\frac{e-1+1}{1-e}\right)=-\frac1{e-1} \text{exp}\left(\frac{1}{1-e}\right) $$
Proof of the equation:
Note that $\text{(strange W)}_{mn}=W_m-W_n$(I spent a while to figure this out)
Note that the inverse of $W$ is a rather simple function. Then we're looking for a value $x$ such that $$x e^x = (x+1)e^{x+1}$$ and this directly gives $$x = \frac{e}{1-e}.$$ So for $W$ we get the argument $$ \frac{e}{1-e}\exp\left(\frac{e}{1-e}\right) = \frac1{1-e}\exp\left(\frac{1}{1-e}\right).$$
