Why didn't they simplify $x^y=y^x$ to $x=y$? Solving $x^y = y^x$ analytically in terms of the Lambert $W$ function
This "solution" for $x^y=y^x$ should simplify to $y=x$, but for some reason no pointed that out in the OP.
According to the stack exchange, the answer is $y= \frac{-xW(-\frac{ln(x)}{x})}{ln(x)}$. However, the term $\frac{-ln(x)}{x}$ itself can be rewritten as 
$$\frac{-ln(x)}{x}=-ln(x)e^{-ln(x)}$$
Therefore, the productlog of that expression should simplify as follows,
$y= \frac{-xW(-\frac{ln(x)}{x})}{ln(x)}, \ \ \ \ \ $ $y= \frac{-xW(-ln(x)e^{-ln(x)})}{ln(x)}, \ \ \ \ \ $ $y=\frac{-x(-ln(x))}{ln(x)}=x$
Did this simplification just slip past everyone or is there something wrong about my algebra? 
 A: The solution is:
$$y = -\frac{x W\left(-\frac{\log (x)}{x}\right)}{\log (x)}$$
which has the following form:

Clearly there are solutions other than $x = y$.  Indeed, we see that for $y=2$ we can have $x=2$ or $x=4$ (intersection between blue and red dashed line).
A: The Lambert $W$ function is not single-valued for negative arguments.  

Using your "simplification" forces use of the lower branch, $W \leq -1$ when you assume $W^{-1}(-\ln x)$ only equals $-\ln (x) \mathrm{e}^{- \ln x}$.  (The same thing happens when you assume the only square root of $3^2$ is $3$ or the only arcsine of $1$ is $-3\pi/2$.)  You get two values from $W^{-1}(-\ln x)$ having the same algebraic form, but one has $0 < x \leq \mathrm{e}$ and one has $x > \mathrm{e}$.  ("$3^2$" and "$(-3)^2$" have the same algebraic form, "$x^2$", but one has $x>0$ and one has $x < 0$.)
This is indicated explicitly in the identities at the Lambert $W$ function article on the English Wikipedia.
Edit: Got myself turned around with too many minus signs.  I originally claimed the $x=y$ solutions were on $W \geq -1$, but this is backwards.  It is corrected above.
