Find the value of $\sqrt{10\sqrt{10\sqrt{10...}}}$

Denote the given problem as $x$, then \begin{align} x&=10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{\cdots}}}}}\\ &=10\cdot10^{\large\frac{1}{2}}\cdot10^{\large\frac{1}{4}}\cdot10^{\large\frac{1}{8}}\cdot10^{\large\frac{1}{16}}\cdots\\ &=10^{\large1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots}\\ &=10^{\large y} \end{align} where $y$ is an infinite geometric series in which its value is \begin{align} y &=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots\\ &=\frac{1}{1-\frac{1}{2}}\\ &=2 \end{align} Therefore \begin{equation} x=10^{\large 2}=100 \end{equation}

If $$x=10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{...}}}}}$$ then it is true that $x=10\sqrt x$.

It is not true that if $x=10\sqrt x$, then $$x=10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{...}}}}}.$$

$0$ is indeed one solution, but it is not a valid one.

Here the question is (apparently) about the sequence defined recursively by $$ \begin{aligned} a_1&=10,\\ a_{n+1}&=10\sqrt{a_n}\quad \text{for all positive integers $n$.} \end{aligned} $$ It often happens that students embark on a quest of finding the value of the limit of a sequence before ascertaining that what they try to calculate actually exists. Here we should first satisfy ourselves that $\lim_{n\to\infty}a_n$ actually exists as a real number. A convenient tool for this is the theorem telling us that a bounded increasing sequence converges towards some limit $A$. Here it is easy to prove by induction that $a_{n+1}>a_n$ for all $n$, and also that $a_n<100$ for all $n$. A part of that theorem then says that $a_1\le A\le 100$. After all, all the members of the sequence are in the closed interval $[10,100]$ so their limit cannot be outside this interval either. The recurrence relation (and continuity of the square root function in this interval) then give us the equation $A=10\sqrt{A}$ allowing us to deduce that either $A=0$ or $A=100$. The former solution, however, does not belong to this interval and can be discarded as an alternative.

Your equation having another solution is just a coincidence. In this case as well as with the recurrence formula $a_{n+1}=\sqrt{6+5a_n}$.

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Compare with the following other common misuse of limits: $$ S=1-1+1-1+1-1+\cdots $$ Some are want for rewriting this as $$ S=1-(1-1+1-1+1-\cdots)=1-S, $$ solving the equation $S=1-S$, and concluding that $S=1/2$.

Where's the mistake? It is in the first line, where the student treats the sum of this series as if it has a value, i.e. as if it converges, by the act of calling it $S$. The moral: Beware of naming things that don't necessarily exist :-)