[tex]x^n = y \Leftrightarrow \sqrt[n]{x^n} = x = \sqrt[n]{y}[/tex], no?
It's because [tex]\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}[/tex] is only true for [tex]a \in \Bbb{R}^+ \cup \{0\}[/tex] and [tex]b \in \Bbb{R}^+ \cup \{0\}[/tex] or something like that. That's the "trick" or whatever... There is another "proof" that says that any integer a equals any other integer b, I just can't remember that one, but there is of course an error in that one as well.