They gave me 0 points for this problem. I think it's unfair. What do you think of this proof, is it correct?
$\lim\limits_{n\to\infty} \underset{2n\text{ roots }}{\underbrace{\sqrt{20\sqrt[3]{15\sqrt{20\sqrt[3]{15}\dots\sqrt{20\sqrt[3]{15}}}}}}}$
$x_n=\underset{2n\text{ roots }}{\underbrace{\sqrt{20\sqrt[3]{15\sqrt{20\sqrt[3]{15}\dots\sqrt{20\sqrt[3]{15}}}}}}}$
$x_1=\sqrt{20\sqrt[3]{15}}$
$x_{n+1}=\sqrt{20\sqrt[3]{15x_n}}=\sqrt[6]{20^3\cdot15x_n}=\sqrt[6]{120000x_n}$
Let's prove that $x_n<\sqrt[5]{120000}$
Basis
$n=1$; $x_1=\sqrt{20\sqrt[3]{15}}=\sqrt[6]{120000}<\sqrt[5]{120000}$Inductive step
given that $x_n<\sqrt[5]{120000}$ let's prove that $x_{n+1}<\sqrt[5]{120000}$
$x_{n+1}=\sqrt[6]{120000x_n}$
$x_{n+1}^5=\sqrt[6]{120000^5x_n^5} < \sqrt[6]{120000^5\cdot120000}=\sqrt[6]{120000^6}=120000$
then
$x_{n+1}<\sqrt[5]{120000}$
Q.E.D.Then
$\frac{x_{n+1}}{x_n} = \frac{\sqrt[6]{120000x_n}}{x_n} = \frac{\sqrt[6]{120000}}{x_n^{5/6}} = \sqrt[6]{\frac{120000}{x^5}} > \sqrt[6]{\frac{120000}{120000}} =1$
$x_{n+1}>x_n$
$\{x_n\}$ is bound from above and increasing, then it has a limit
Let's denote it as $x$, then
$\lim\limits_{n\to\infty} x_{n+1} = \lim\limits_{n\to\infty} \sqrt[6]{120000}x_n$
$x=\sqrt[6]{120000x}$
$x^6=120000x$
$x_n>1$ thus $x\ge1$ than $x\ne0$
$x=\sqrt[5]{120000}$
$\lim\limits_{n\to\infty} \underset{2n\text{ roots }}{\underbrace{\sqrt{20\sqrt[3]{15\sqrt{20\sqrt[3]{15}\dots\sqrt{20\sqrt[3]{15}}}}}}}= \sqrt[5]{120000}$
Your proof in general looks absolutely correct (and fairly standard) to me, and if partial credit is possible I would not consider this to be worth $0$ points.
I do see two points where there might be some slight complaint: first you might want to specifically reference the monotone convergence theorem, and second you should justify your claim that $x_n>1$ in the third to last line (this statement can be easily proven via induction), but these are very minor qualms.