How to solve $x^x = x?$

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Solve: $x^x = x\quad (*)$

I can only solve it when $x > 0$

$(*)\Leftrightarrow \ln(x^x) = \ln x$

$\Leftrightarrow x\ln x - \ln x = 0$

$\Leftrightarrow \ln x(x -1) = 0$

Then $x = 1$

How can I solve it with $x \leq -1$ ?

If it's possible, please show me how to find its domain also.

Thank you for helping!!!