if $x_n=\sqrt{2+\sqrt[3]{3+\sqrt[4]{4+...\sqrt[n]{n}}}}$ prove that for $n\ge 2$ $$x_{n+1}-x_n<\frac{1}{n!}$$.
I think induction works best here. The case when $n=2$ is easy as $x_3-x_2=\sqrt{2+\sqrt[3]{3}}-\sqrt{2}=0.44<\frac{1}{2!}$.
However i am not able to proceed further ,the complex radicals are creating a lot of trouble
Hint: use the fact that $$\sqrt{y}-\sqrt{x}=\frac{y-x}{\sqrt{y}+\sqrt{x}}$$ For example $$x_{n+1}-x_n=\frac{\sqrt[3]{3+\sqrt[4]{4+...\sqrt[n+1]{n+1}}}-\sqrt[3]{3+\sqrt[4]{4+...\sqrt[n]{n}}}}{\sqrt{2+\sqrt[3]{3+\sqrt[4]{4+...\sqrt[n+1]{n+1}}}}+\sqrt{2+\sqrt[3]{3+\sqrt[4]{4+...\sqrt[n]{n}}}}}<\\ \frac{\sqrt[3]{3+\sqrt[4]{4+...\sqrt[n+1]{n+1}}}-\sqrt[3]{3+\sqrt[4]{4+...\sqrt[n]{n}}}}{2}<...$$ Next $$\sqrt[3]{y}-\sqrt[3]{x}=\frac{y-x}{\sqrt[3]{y^2}+\sqrt[3]{yx}+\sqrt[3]{x}}$$ leading to $$...<\frac{1}{2}\frac{\sqrt[4]{4+...\sqrt[n+1]{n+1}}-\sqrt[4]{4+...\sqrt[n]{n}}}{3}$$ And so on ...
General idea comes from the following identity $$y^n-x^n=(y-x)(y^{n-1}+y^{n-2}x+y^{n-3}x^2+...+yx^{n-2}+x^{n-1})$$ Replace $y\to \sqrt[n]{y}$, $x\to \sqrt[n]{x}$ and consider that both $y>1$ and $x>1$. As a result $$\sqrt[n]{y}-\sqrt[n]{x}<\frac{y-x}{n}$$