Evaluate the general infinite square root $$x = \sqrt{n\sqrt{n\sqrt{n}} \cdots}$$

I see that:

$$x = \sqrt{nx}$$

$$x^2 -nx = 0$$

Them:

$$x(x - n) = 0 \implies x \in \{0, n\}$$

How should I reject the $x = 0$ solution?  (any level proof is fine, analysis, calculus etc...)
We have $$x = n^{1/2}n^{1/4}n^{1/8}\cdots = n^{\sum_{k = 1}^\infty \left(\frac{1}{2}\right)^k} = n^{\frac{1/2}{1 - 1/2}} = n$$
Assuming that $x=0$ is also a solution with $n \neq 0$. Then you would have the relation $0 = \sqrt{n \sqrt{n \sqrt {n ...}}}$. Now one can take the square on both sides and it follows that either $n$ is zero or the iterated square root is zero. When $n=0$ you are done; else you continue proving the rejection of 0 by squaring again on both sides and compare solutions. Since this procedure is repeadet infinitely many, one concludes that $n=0$ which is a contradiction.