The proof combines two ingredients:

Ingredient 1: By Galois's criterion, a polynomial $f$ over $\mathbb{Q}$ is solvable by radicals iff $\mathrm{Gal}(f) = \mathrm{Gal}(K/\mathbb{Q})$ is a solvable group (where $K$ is the splitting field of $f$).

Ingredient 2: $S_n$ is not solvable for $n \geq 5$ (since $A_n$ is simple for $n \geq 5$).

Ingredient 3: There exist polynomials of degree $n$ over $\mathbb{Q}$ with $\mathrm{Gal}(f) = S_n$.

For Ingredient 3 with $n = 5$: the polynomial $f(x) = x^5 - 4x + 2$ has the following properties:

• It is irreducible over $\mathbb{Q}$ (by Eisenstein with $p = 2$).
• It has exactly 3 real roots (check by calculus: $f'(x) = 5x^4 - 4$, two critical points, sign analysis).
• Therefore $\mathrm{Gal}(f)$ acts transitively on 5 roots and contains a transposition (complex conjugation swaps the two non-real roots). A transitive subgroup of $S_5$ containing a transposition is $S_5$ itself (since $5$ is prime, transitivity gives a 5-cycle, and a 5-cycle plus a transposition generate $S_5$).

Combining: $\mathrm{Gal}(f) = S_5$ is not solvable, so $f$ is not solvable by radicals. $\blacksquare$