[step: Galois group of the general polynomial is $S_n$] Let $f(t) = t^n - e_1 t^{n-1} + e_2 t^{n-2} - \cdots + (-1)^n e_n$ be the general polynomial of degree $n$ over the field $F = \mathbb{Q}(e_1, \dots, e_n)$, where $e_1, \dots, e_n$ are independent indeterminates. Its roots $\alpha_1, \dots, \alpha_n$ in the splitting field $K = \mathbb{Q}(\alpha_1, \dots, \alpha_n)$ satisfy $e_k = e_k(\alpha_1, \dots, \alpha_n)$, the $k$-th elementary symmetric polynomial. By the fundamental theorem of symmetric polynomials, every element of $F$ fixed by all permutations of the roots already lies in $\mathbb{Q}(e_1, \dots, e_n) = F$, so the fixed field of $S_n$ acting on $K$ is exactly $F$. The extension $K/F$ has degree $[K : F] = n!$ because the $\alpha_i$ are algebraically independent over $\mathbb{Q}$ except for the symmetric relations, and the Galois correspondence gives \begin{align*} \operatorname{Gal}(K/F) \;\cong\; S_n. \end{align*} [/step] [step: $S_n$ is not solvable for $n \ge 5$] [guided] A group $G$ is solvable when its derived series $G \rhd G' \rhd G'' \rhd \cdots$ terminates at $\{e\}$. For the symmetric group the first derived subgroup is $S_n' = A_n$, since $S_n / A_n \cong \mathbb{Z}/2\mathbb{Z}$ is abelian. The derived series therefore stalls at $A_n$ precisely when $A_n' = A_n$, which happens whenever $A_n$ is non-abelian simple. For $n \ge 5$ the alternating group $A_n$ is simple. To see this, one shows that every non-trivial normal subgroup $N \trianglelefteq A_n$ contains a $3$-cycle: if $\sigma \in N$ is not a $3$-cycle, form the commutator $[\sigma, \tau]$ with a suitably chosen $\tau \in A_n$ to produce an element of $N$ with strictly smaller support; iterate until a $3$-cycle appears. Since all $3$-cycles are conjugate inside $A_n$ (for $n \ge 5$ every conjugation can be performed by an even permutation), $N$ contains every $3$-cycle, hence $N = A_n$. Because $A_n$ is simple and non-abelian, $A_n' = A_n$, so the derived series of $S_n$ is \begin{align*} S_n \;\rhd\; A_n \;\rhd\; A_n \;\rhd\; A_n \;\rhd\; \cdots \end{align*} and never reaches $\{e\}$. Therefore $S_n$ is not solvable for every $n \ge 5$. [/guided] [/step] [step: The solvability theorem forces the conclusion] Galois's solvability theorem states: a polynomial $f \in F[t]$ is solvable by radicals over $F$ if and only if $\operatorname{Gal}(f/F)$ is a solvable group. Since $\operatorname{Gal}(K/F) \cong S_n$ and $S_n$ is not solvable for $n \ge 5$, no formula expressing the roots of the general degree-$n$ polynomial in terms of $e_1, \dots, e_n$ by means of addition, subtraction, multiplication, division, and extraction of $k$-th roots can exist. $\blacksquare$ [/step] [step: Concrete witness — $f(t) = t^5 - 4t + 2$ over $\mathbb{Q}$] [guided] We exhibit a specific quintic with Galois group $S_5$. **Irreducibility.** The polynomial $f(t) = t^5 - 4t + 2$ is Eisenstein at $p = 2$: the constant term $2$ is divisible by $2$ but not by $4$, and the non-leading coefficient $-4$ is divisible by $2$, while the leading coefficient $1$ is not. By Eisenstein's criterion $f$ is irreducible over $\mathbb{Q}$, so $\operatorname{Gal}(f/\mathbb{Q})$ acts transitively on the five roots and embeds into $S_5$ as a transitive subgroup of order divisible by $5$. In particular it contains a $5$-cycle. **Root signature.** We compute $f'(t) = 5t^4 - 4$, which vanishes at $t = \pm (4/5)^{1/4}$. Evaluating: \begin{align*} f\!\bigl(-(4/5)^{1/4}\bigr) &> 0, \qquad f\!\bigl((4/5)^{1/4}\bigr) < 0, \end{align*} together with $f(-\infty) = -\infty$ and $f(+\infty) = +\infty$, gives exactly three real roots and one pair of complex conjugate roots. **Transposition from complex conjugation.** Complex conjugation $\sigma \in \operatorname{Gal}(f/\mathbb{Q})$ fixes the three real roots and swaps the two complex roots, so $\sigma$ is a transposition in $S_5$. **Generation.** A transposition and a $5$-cycle together generate the full symmetric group $S_5$ (classical fact: a transposition and a $p$-cycle generate $S_p$ for $p$ prime). Therefore \begin{align*} \operatorname{Gal}(t^5 - 4t + 2\;/\;\mathbb{Q}) \;\cong\; S_5, \end{align*} and this particular quintic is not solvable by radicals. $\blacksquare$ [/guided] [/step]