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The polynomial $X^3+X+1$ has no root in ${\mathbb{F}}_2$, so it is an irreducible polynomial. The quotient ring of ${\mathbb{F}}_2[X]$ by the ideal it generates is therefore a field with eight elements. The seven nonzero elements form a group of order seven, so each satisfies $x^7=1$ by Lagrange's theorem. Thus this field splits $X^7-1$. A root different from one has order seven; it cannot lie in ${\mathbb{F}}_2$ and generates this degree-three field extension, since three is prime and the tower law applies. Therefore

An element $\alpha$ is an algebraic element over $F$ if some nonzero polynomial in $F[X]$ vanishes at $\alpha$. A field extension is an algebraic extension if each of its elements is an algebraic element over $F$. If $[K:F]=d<\infty$, the $d+1$ vectors $1,\alpha ,\dots ,{\alpha }^d$ are linearly dependent, producing such a polynomial. Thus every finite field extension is an algebraic extension.

The converse fails for the field of algebraic numbers $\overline{\mathbb{Q}}$. It is an algebraic extension of $\mathbb{Q}$ by definition, but it contains primitive $p$th roots of unity of degree $p-1$ for arbitrarily large primes $p$, by the permitted irreducibility of the cyclotomic polynomial. The tower law would make every such degree at most $[\overline{\mathbb{Q}}:\mathbb{Q}]$ if the latter were finite, a contradiction.

A regular heptagon is a regular polygon with seven sides. Its unit-circle vertices have complex coordinates $e^{2\pi ik/7}$ for $k=0,\dots ,6$. It cannot be constructed from a unit segment by a straightedge and compass construction: the first coordinate $\cos (2\pi /7)$ has degree three over $\mathbb{Q}$, since the seventh cyclotomic field has degree six and degree two over its real subfield. Every finite constructible coordinate extension has power-of-two degree, which cannot contain a degree-three subextension by the tower law.