In class today I was learning about the old factorization of $x^2-1$ to be $(x+1)(x-1)$.
I was thinking that $x-1$ could be factored to $(\sqrt[2]{x}+1)(\sqrt[2]{x}-1)$.
And in turn, $\sqrt[2]{x}-1$ could then be factored into $(\sqrt[4]{x}+1)(\sqrt[4]{x}-1)$.
I thought that possibly this process could continue forever, with the same solution of $x^2-1$.
What I wanted to ask is whether or not $\prod\limits_{n=1}^{\infty}(\sqrt[2^n]{x}+1)$ would truly be equal to $x-1$.