TITLE: Solution of $W_0(x)-W_{-1}(x)=1$
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: 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 $$ \frac{e}{1-e}\exp\left(\frac{e}{1-e}\right) = \frac1{1-e}\exp\left(\frac{1}{1-e}\right).$$