Lambert W function multiplied by a constant Consider the equation
$$y=x e^x.$$
Its real solution is given by
$$x=W(y),$$
where $W$ is the Lambert W function (or product log).
Can the function
$$f(x) = W\left(-\frac{1}{r}xe^x\right)$$
be written in simple terms of $x$?
What I've tries so for is writing this as a power series expansion
$$W(x)=\sum_{n=1}^{\infty}\frac{(-1)^n n^{n-2}}{(n-1)!}x^n$$
therefore with the constant I've introduced
$$
\begin{split}
W\left(-\frac{1}{r}x\right) & =\sum_{n=1}^{\infty}\frac{(-1)^{n-1} n^{n-2}}{(n-1)!}\left(-\frac{x}{r}\right)^n\\
& =\sum_{n=1}^{\infty}-\frac{(-1)^{n-1}(-1)^{n-1} n^{n-2}}{(n-1)!}\left(\frac{x}{r}\right)^n \\
& =\sum_{n=1}^{\infty}-\frac{n^{n-2}}{(n-1)!}\left(\frac{x}{r}\right)^n
\end{split}
$$
but got stuck here.
I've also tried guessing many functions in mathematica that yeilded no result.
As I see it now the problem is not solvable
 A: If what you want is a series expansion in powers of $x$ and $t:=1/r$, we can write
$$f(x,t)=W\big(- {txe^x}\big)=-\sum_{n=1}^\infty  {q_n(t)} \,x^n=-\bigg[tx+\frac{2t+2t^2}{2!}\, {x^2} +\frac{3t+12t^2+9t^3}{3!}\,x^3+\dots\bigg]$$
The coefficient $q_n$ is a  polynomial  of degree $n$, precisely
$$q_n(t)=\frac{1}{n!}(tD)^{n-1}(1+t)^n$$
where $D$ denotes differentiation in $t$.  
A: Maple did the following.  (Of course once you see it you can do it yourself...)
$$
y = W(-\frac{1}{r}xe^x)
$$
solve for $x$; result
$$
x = W(-rye^y)
$$

Probably even your original assertion needs adjustment.  We may think that
$$
W(xe^x) = x
$$
but perhaps not.  Here is the graph for $W_0(xe^x)$

Only part of that is $y=x$.   
For more explanation, see this, where $W_0(xe^x)$ is in red and $W_{-1}(xe^x)$ is in blue:

A: As mentioned by Gerald Edgar, you seek to simplify 
$$y = W(-r^{-1}xe^x)\Rightarrow x=W(-rye^y),$$
for $r>0$. Here is a series expansion in powers of $r$,
$$x=-\sum_{n=1}^\infty n^{n-1}\frac{1}{n!}(rye^y)^n.$$
The plot compares the exact result for $x$ (blue curve) with the series expansion up to $n=8$ (gold) at $y=-1$, as a function of $r$. (I take $y<0$ because the OP says the interest is in $x>0$.) The series expansion remains quite accurate even for $r$ approaching unity.

