Evaluating $\sqrt{7\sqrt{7\sqrt{7\sqrt{7\,\cdots}}}}$. How can we know $0$ is an extraneous solution? The task is to find the the value of this expression:
$$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\,\cdots}}}}$$
We can assume
$$\sqrt{7\sqrt{7\sqrt{7\sqrt{7....}}}} = x$$
Then we replace the nested part with $x$
$$\sqrt{7x} = x$$
Then square both sides and solve the equation
$$\begin{align}
7x &= x^2 \\[4pt]
x^2 - 7x &= 0 \\[4pt]
x(x - 7) &= 0 \\[1em]
x_1 = 7,\quad& x_2 = 0
\end{align}$$
The problem is I assume $0$ is not a valid solution, but I can't explain why. It almost makes sense it could be $0$ as the numbers get progressively smaller.
Can someone explain how can we reject the solution $0$?
Edit: I think I understand it better now. But what I'm most confused about is why we get the invalid solutions here in the first place.
Could it be that the problem is in the beginning where we could write the equation in different ways, like $\sqrt{7\sqrt{7x}} = x ?$ This way we would get 4 different solutions.

Extraneous solutions are usually a result of an algebraic manipulation of an equation. In this case, by squaring the equation, you are forcing  it to have “two” solutions. This is the reason why it pays to check if solutions make sense.
