Is this a sufficient flow of logic?
Consider a fifth degree polynomial that has a formulaic solution. Then we have a radical extension with a string of subgroups of the Galois group which each have a quotient that is abelian. Within the string, the subgroups must have all 3-cycles in $S_5$
Suppose there exists a subgroup with all 3 cycles. Since its quotient is abeleian, two 3-cycles and multiply by there inverses we can argue all 3-cycles are in the subgroup. However there are no cycles in the identity, so this is a contradiction. Thus we have no formula to solve this quintic polynomial.