xxxx*x=x equation

Given the equation
$$x x x x x = x$$
Obviously:

x0 = 0

next,
transform
$$\frac{1}{x^{4}} = 1$$
Because equation degree is equal to = -4 - contains the even number -4 in the numerator, then
the equation has two real roots.
Get the root -4-th degree of the equation sides:
We get:
$$\frac{1}{\sqrt[4]{\frac{1}{x^{4}}}} = \frac{1}{\sqrt[4]{1}}$$
$$\frac{1}{\sqrt[4]{\frac{1}{x^{4}}}} = \left(-1\right) \frac{1}{\sqrt[4]{1}}$$
or
$$x = 1$$
$$x = -1$$
We get the answer: x = 1
We get the answer: x = -1
or
$$x_{1} = -1$$
$$x_{2} = 1$$

All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$\frac{1}{z^{4}} = 1$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$\frac{e^{- 4 i p}}{r^{4}} = 1$$
where
$$r = 1$$
- the magnitude of the complex number
Substitute r:
$$e^{- 4 i p} = 1$$
Using Euler’s formula, we find roots for p
$$- i \sin{\left(4 p \right)} + \cos{\left(4 p \right)} = 1$$
so
$$\cos{\left(4 p \right)} = 1$$
and
$$- \sin{\left(4 p \right)} = 0$$
then
$$p = - \frac{\pi N}{2}$$
where N=0,1,2,3,...
Looping through the values of N and substituting p into the formula for z
Consequently, the solution will be for z:
$$z_{1} = -1$$
$$z_{2} = 1$$
$$z_{3} = - i$$
$$z_{4} = i$$
do backward replacement
$$z = x$$
$$x = z$$

The final answer:

x0 = 0

$$x_{1} = -1$$
$$x_{2} = 1$$
$$x_{3} = - i$$
$$x_{4} = i$$