smc

We can write $\sqrt[3]{10}$ as $10^{\frac{1}{3}}$. Similarly, $\sqrt[3]{\sqrt[3]{10}}=(10^{\frac{1}{3}}{)}^{\frac{1}{3}}=10^{\frac{1}{3^2}}$. By continuing this, we get the form $$10^{\frac{1}{3}}\cdot 10^{\frac{1}{3^2}}\cdot 10^{\frac{1}{3^3}}\cdots ,$$ which is $$10^{\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots }.$$ Using the formula for an infinite geometric series $S=\frac{a}{1-r}$, we get $$\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots =\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{1}{2}.$$ Thus, our answer is $10^{\frac{1}{2}}=\boxed{\sqrt{10}}$.