Solution: We need to show that $[\mathbb{F} : \mathbb{Q} ]$ is infinite. Let us assume that it is finite and $[\mathbb{F} : \mathbb{Q} ] = n$. Choose a prime number $p$ such that $p > n + 2$. Then the polynomial \[ x^p - 1 = (x - 1) \left( x^{p-1} + \dots + x + 1 \right) \implies \frac{x^p - 1}{x - 1} = x^{p-1} + \dots + x + 1. \] By Eisenstein criteria (for irreducibility of a polynomial), this is an irreducible polynomial as the polynomial \begin{align*} \frac{(x + 1)^p - 1}{(x + 1) - 1} = x^{p-1} + p x^{p-2} + \dots p x + p \end{align*} is irreducible. Since $\mathbb{F} $ is algebraic, every $p^{\text{th} }$ root of unity is in $\mathbb{F} $. Hence for the prime $p$, there exists an element $\alpha _p$ algebraic over $\mathbb{Q} $ and $[\mathbb{Q} (\alpha _p): \mathbb{Q} ] \geq p-1$. Thus, \begin{align*} n = [\mathbb{F} : Q] \geq [\mathbb{Q} (\alpha _p) : \mathbb{Q} ] \geq p-1 > n + 1, \end{align*} a contradiction. Thus, $[\mathbb{F} : \mathbb{Q} ]$ is infinite.