let $x = (((((2^2)^2)^2)^2)^2)^\cdots$
Therefore, $x^2 = ((((((2^2)^2)^2)^2)^2)^2)^\cdots$
But this is the same expression.
Therefore $x = x^2$
Therefore, $x^2 - x = 0$
Therefore $x(x-1) = 0$
Therefore $x = 0$, or $x = 1.$
Where is the error in this argument? I assume there is an error as repeated power of $2$'s should make the number larger and larger.
On
Stating $x^2=$ an expression implies that the expression is well defined. When it is of infinite length, it needs to converge to something. In this case, it does not.
On
The mistake is that you are treating $((2^2)^2)^2)\dots$ like it's a number. "..." actually means something, and this argument does not acknowledge its meaning. To understand what "..." is usually taken to mean, consider the number $0.333...$. The number $0.333...$ can be written as$$0.333...=\frac{3}{10}+\frac{3}{100}+\frac{3}{1000}+\dots$$ And what is actually meant by this is that the number which is represented symbolically by $0.333...$ is the limit of the sequence $$\{\frac{3}{10},\frac{33}{100},\frac{333}{1000},\dots\}$$It can be shown using tools from calculus that this particular sequence converges, or in other words that the limit exists (and it just so happes to be $\frac{1}{3}$). But if it is understood that "..." refers to the limit a convergent sequence, then if that limit doesn't exist, it makes no sense to use "...", at least not and treat it as a number. If you wanted to abuse notation, but hopefully in a slightly less bad way, you could rewrite this as
$$\infty=((2^2)^2)^2...$$
Now it's not clear what it even means to square the left hand side of the above equation, but that only highlights how little sense it makes to square the right hand side and treat the result as a real number.
On
When you solve an equation like x^2 = x for x, you usually implicitly specify a field in which the solution lies. Usually the implicit field is the real numbers, but this choice is arbitrary and context dependent. You might choose all the rational numbers AND the square root of two (and all of it's multiples, and it's multiplicative inverse), for example.
Infinity is usually not in this implicit field. As such, your candidate solution, 2^2^2^2 ..., which does not to converge in the reals, or the complex numbers, is not a valid solution.
If infinity is in the field (you could, for example, be in the projective plane), you have to set up rules for how to treat it. As you have demonstrated, these rules would have to maintain that infinity is a solution to x^2 - x = 0.
$\infty$ is also a solution of $x^2 = x$, since $\infty^2 = \infty$. It is the correct solution in this case.