It's almost trivial!

Firstly, note that the $LHS$ is not defined $\forall x \in [-1, 0]$ .

For $x>0$ , the $LHS$ is continuously decreasing towards $e>RHS$ .

For $x<-1$ , the $LHS$ is still continuously monotonous, so there exists at most one solution.

Now, we note that

$\begin{aligned} \left ( 1 + \dfrac{1}{1999} \right )^{1999} &= \left (\dfrac{2000}{1999} \right )^{1999}\\ \left ( 1 + \dfrac{1}{x} \right )^{x+1} &= \left (\dfrac{1999}{2000} \right )^{-1999}\\ &= \left (1+\dfrac{1}{-2000} \right )^{-2000+1}\\ \end{aligned}$

Thus, $x=-2000$ , and this is the only solution.