All square roots can be expressed as a regular Continued Fraction. For example: $$\newcommand{\contfrac}{\raise{-0.5ex}\mathop{\Large\mathrm{K}}}\sqrt{2}+1=2+\contfrac_{n = 0}^\infty \frac{1}{2}$$
It is easy to see that comparable regular Continued Fractions for higher order roots are impossible. But at least some higher roots can actually be expressed as Continued Fractions, following simple patterns. For example:
$$4\sqrt[3]{4}-1=7+\contfrac_{n = 0}^\infty \frac{-4\cdot\frac{3n+2}{n+1}}{7} =7+\cfrac{-4\cdot\cfrac{2}{1}}{7+\cfrac{-4\cdot\cfrac{5}{2}}{7+\cfrac{-4\cdot\cfrac{8}{3}}{7+\cfrac{-4\cdot\cfrac{11}{4}}{7+\ddots}}}}$$ $$6\sqrt[5]{6}-1=11+\contfrac_{n = 0}^\infty \frac{-6\cdot\frac{5n+4}{n+1}}{11}$$
$$(1+p)\sqrt[p]{p+1}-1=1+2p+\contfrac_{n = 0}^\infty \frac{-(1+p)\cdot\frac{(n+1)p-1}{n+1}}{1+2p}$$ I have proven these CF's by applying Euler's Differential Method for Continued Fractions.
The question remains: is there a way to express an arbitrary higher order root like $\sqrt[3]{5}$ as a "patterned" Continued Fraction?
You can definitely express higher order roots as Continued Fractions. However my little experience with the General Continued Fractions proved most of them to be trash because almost always they converge to the target with a snail's pace.
For higher order roots as Continued Fractions, using the Simple Continued Fraction approach is the most efficient and also the simplest way. It's like every discovered coefficient takes up at least a decimal place in the result. As an example, when I test $2190^\frac{1}{7}$ in my code, only 50 SCF coefficients converge to 72 correct decimal places.
However it is kind of a little convoluted and also a long topic to explain here since they involve calculations like translating, inverting and finding the positive root's lower bound of $n^{th}$ order polynomials... $n$ being the higher order that you mention.
If you would like to know more then perhaps an earlier post of mine may be of your interest.