I can give a very fast and incomplete explanation, but you must trust me.
In the complex numbers, the vertices of a pentagon are z^5-1=0. You can factorize it as (z^4+z^3+z^2+z+1)*(z-1)=0. The hard part is solving z^4+z^3+z^2+z+1=0.
Now that equation can't be factorized, and has degree 4. It's important that the solutions have a property that is related to the degree of the equation so they have a property that is 4.
With a compass and a straightedge you can solve only equations of degree 2, that is like taking a square root. If you repeat the process you can solve (some) equations of degree 4. So after a few tricks, you can solve the equation and draw the pentagon.
For 17, the equation is z^16+z^15+...+z+1=0. So the property is 16 and you must use the square root a few times. Each time the solutions double their property, so you get 1 -> 2 -> 4 -> 8 -> 16. Near the bottom of the article is the formula, and it's possible to see a lot of nested and repeated square roots.
For 7, the equation is z^6+z^5+...+z+1=0. So the property of the solutions is 6. With the square root you can only double the property, so you get 1 -> 2 -> 4 -> 8 -> 16 -> 32 ... but you can never get a solution which has a property equal to 6.
(There are more technicals details. You can solve some equations of degree 16, for example to draw the 17agon, but you can't solve every one of them.)
For example with 9, you can factorize z^9-1=0 as (z^6+z^3+1)*(z^2+z+1)*(z-1)=0, and now the property to calculate is 6*2 instead of 8, so it's not a power of 2 and the polygon is impossible to construct.
In the complex numbers, the vertices of a pentagon are z^5-1=0. You can factorize it as (z^4+z^3+z^2+z+1)*(z-1)=0. The hard part is solving z^4+z^3+z^2+z+1=0.
Now that equation can't be factorized, and has degree 4. It's important that the solutions have a property that is related to the degree of the equation so they have a property that is 4.
With a compass and a straightedge you can solve only equations of degree 2, that is like taking a square root. If you repeat the process you can solve (some) equations of degree 4. So after a few tricks, you can solve the equation and draw the pentagon.
For 17, the equation is z^16+z^15+...+z+1=0. So the property is 16 and you must use the square root a few times. Each time the solutions double their property, so you get 1 -> 2 -> 4 -> 8 -> 16. Near the bottom of the article is the formula, and it's possible to see a lot of nested and repeated square roots.
For 7, the equation is z^6+z^5+...+z+1=0. So the property of the solutions is 6. With the square root you can only double the property, so you get 1 -> 2 -> 4 -> 8 -> 16 -> 32 ... but you can never get a solution which has a property equal to 6.
(There are more technicals details. You can solve some equations of degree 16, for example to draw the 17agon, but you can't solve every one of them.)