This is the product $3^{\frac{1}{2}}\cdot5^{\frac{1}{4}}\cdot3^{\frac{1}{8}}\cdot\ldots$ $=3^{\frac{1}{2}+\frac{1}{8}+\ldots}\cdot5^{\frac{1}{4}+\frac{1}{16}+\ldots}$ $=3^{\frac{2}{3}}\cdot5^{\frac{1}{3}}=\sqrt[3]{45}$.

another solution: denote ${{\sqrt{3\sqrt{5\sqrt{3...}}}}=x, \sqrt{5\sqrt{3\sqrt{5...}}}}=y$. We have $\sqrt{3y}=x, \sqrt{5x}=y$. From this system we have $x=\sqrt[3]{45}$.Ok

Show that:\[\sqrt{2\sqrt[3]{3\sqrt[4]{4\cdots\sqrt[n]{n}}}}<2.\]

Let $$ x= \sqrt{3\sqrt{5\sqrt{3\sqrt{5...}}}}$$ Squaring both sides, we have $$x^2=3\sqrt{5x}$$ Squaring again, we have $$x^4=45x$$$$x^4-45x=0$$$$x(x^3-45)=0$$$$x=0 \text{ and } x=\sqrt[3]{45}$$ Since the expression is impossible to be $0$, therefore the answer is $\sqrt[3]{45}.$