I have a method to find the radical expression of $n$-th root of unity but it looks too good to be true (and most probably differs from the standard proof given in introductory galois theory books which I'm reading). Can anyone point out the mistake ?
We induct on $n$. Without loss of generality, assume $n$ is a prime number $p$.
Lemma 1: Let $F$ be a field, and $P \in F[x]$ be an irreducible polynomial with roots $x_1, \cdots, x_n$. Let $g \in F[y_1, \cdots, y_n]$ such that $g(y_1, \cdots, y_n) \neq g(y_{\pi(1)}, \cdots, y_{\pi(n)})$ for any non identity permutation $\pi$. Define $r = f(x_1, \cdots, x_n)$. Then for each $i \in \{1,\cdots, n \}$ there exits $M_i, N_i \in F[x]$ such that $x_i = \frac{M_i(r)}{N_i(r)}$.
Proof: Well known, follows from `` Lagrange's lemma "
Lemma 2: If $(m,n) = 1$ and $\tau, \delta$ are $m$th, $n$th root of unity respectively, then $[\mathbb{Q}(\tau, \delta): \mathbb{Q}] = \phi(m) \phi(n)$
Proof: Let $\sigma$ be a $mn$th root of unity. Then $\mathbb{Q}(\tau, \delta) \subset \mathbb{Q}(\sigma)$ trivially since $\sigma^n = \tau$ and $\sigma^m = \delta$. Since $\gcd(m,n) = 1$, pick $a,b$ such that $am - bn = 1$, and then $\sigma = \frac{\tau^a}{\delta^b}$ so $\mathbb{Q}(\sigma) \subset \mathbb{Q}(\tau, \delta)$. So infact we have $\mathbb{Q}(\sigma) = \mathbb{Q}(\tau, \delta)$. Now $\sigma$ has minimal polynomial $\Phi_{mn}(x)$ which is irreducible over $\mathbb{Q}$ and has degree $\phi(mn) = \phi(m) \phi (n)$, so we get $$ [\mathbb{Q}(\tau, \delta): \mathbb{Q}] = [\mathbb{Q}(\sigma) : \mathbb{Q} ] = \phi(mn) = \phi(m) \phi(n)$$
Lemma 3: Let $g$ be a primitive root of unity modulo $p$, and $\omega$ be a complex $p$th root of unity and $\rho$ be a complex $p-1$th root of unity. Define $K = \mathbb{Q}[\rho]$. Then $z \in K(\omega)$ is fixed under the automorphism $K(\omega) \mapsto K(\omega^g)$ iff $z \in K$
Proof: By the previous theorem, we have $$[K(\omega): K] = \frac{[K(\omega): \mathbb{Q}]}{[K: \mathbb{Q}]} = \frac{\phi(p \cdot (p-1)}{\phi(p-1)} = \phi(p) = p-1$$, so the $\Phi_p(x)$ is irreducible over $K$ also and thus if we have $G(\omega) = a_0 + a_1 \omega + \cdots + a_{p-1} \omega^{p-1} = 0$ then $P(x) | G(x)$ and thus $a_0 = a_1 = \cdots = a_{p-1}$.
Now pick an element $z \in K(\omega)$ which is fixed under that automorphism. We can WLOG write $\displaystyle z = a_0+\sum_{1 \leq i \leq p-1} a_i \omega^{g^i}$, so the automorphism maps $z$ to $z' = a_0+ \displaystyle \sum_{1 \leq i \leq p-1} a_{i} \omega^{g^{i+1}} = \sum_{1 \leq i \leq n-1} a_{i-1} \omega^{g^{i+1}}$. Now $0 = z - z' = \displaystyle \sum_{1 \leq i \leq p-1} \omega^{g^i} (a_i - a_{i+1})$, so by the precedding paragraph we get $a_1 = a_2 = \cdots = a_{p-1} = 0$ which gives the conclusion.
Coming back to the main problem and using the notation of Lemma-3, by induction hypothesis note that all elements of $K$ are expressible by radicals. Now define $f(x_1, \cdots, x_n) := \displaystyle \sum_{1 \leq i \leq p-1} \rho^i x_i \in K[x_1, \cdots, x_n] $. Then $t := f(\omega^g, \omega^{g^2}, \cdots, \omega^{g^{p-1}})$ is invariant under the automorphism $\omega \mapsto \omega^g$ so $t^{p-1} \in K$, so $t$ can be expressed as $\rho^i \sqrt[10]{z}$ for some $z \in K$.
Now note that $f$ satisfies the condition of Lemma-1, so we can find $M_i, N_i \in K[x]$ s.t $\omega = \frac{M_i(t)}{N_i(t)}$. The end.