There are three issues here.

## You did it backwards

You have done what is basically the common high school algebra mistake of doing problems backwards — what you have proven is:

(Assuming that the product satisfies that algebraic identity) if $x$ is a real number solution to $x \cdot x \cdot x \cdot \ldots = 2$, then $x = 1$

In particular, you have completely forgotten to put any work towards showing

If $x=1$, then $x$ is a solution to $x \cdot x \cdot x \cdot \ldots = 2$

## You haven't defined the problem

Multiplication is a binary operation. By induction, you can extend multiplication to any *finite* number of operands.

If you want to talk about multiplying *infinitely* many operands, then the above is insufficient. You have to find some way to make sense of such an operation.

You might do this by considering the "limit of partial products" approach from introductory calculus; e.g.

$$ x \cdot x \cdot x \cdot \ldots := \lim_{n \to \infty} \underbrace{x \cdot x \cdot \ldots \cdot x}_{n \text{ factors}} $$

## You probably misunderstand infinitary operations

I infer this because you write as if:

- Your product has a first factor
- Your product has a last factor
- Your product has infinitely many factors

While it is *possible* to arrange for such a thing, it is unusual and contrary to the common meaning of notation. One who actually understands how to arrange for this would understand the need to actually explain more precisely what they actually mean.

Thus, I imagine you either made a mistake (e.g. you did not mean to put a final term), or simply misunderstand infinitary algebra.