I can make 2 swaps to go 1,2,3,4,5 -> 1,4,3,2,5 -> 1,4,5,2,3. That's an even permutation.
The order (size) of A_n is the number of even permutations of n unique elements. The orders are as follows:
A2: 1
A3: 3
A4: 12
A5: 60
A6: 360
A_n: n! / 2
The "solvability" of a polynomial depends on the the degrees of freedom and number of constraints. Quintic is where the number of degrees of freedom blows up and you can no longer factor out roots in such a way that you can uniquely factor the polynomial. Quadratic is basically trivial. Cubic cannot be directly factored, because you have 3 variables and 2 equations, but you can "factor out" a quadratic that you can solve, and then solve the cubic. Quartic is the same, you end up with 4 variables and 3 unique equations, you can factor out a cubic, which you can solve, etc. Quartic, you have 4 equations to solve and...something like 30 variables (or maybe it's 60. It's some way more than we can factor!).
https://en.wikipedia.org/wiki/Alternating_group
https://math.stackexchange.com/questions/550401/intuitive-re...
Now if you have a general expression built out of extracting nth roots for various n, that might generate some big complicated field extension. But it corresponds to a tower of field extensions which are all just extracting nth roots. Thus the intermediate symmetry groups are still all subsets of Z/n for various n.
The Galois correspondence tells you that the symmetry group of the entire extension, call it G, has a corresponding decomposition: there is an increasing sequence of subgroups 1 < G_1 < … < G_k = G such that each intermediate quotient group G_{i+1}/G_i is a subgroup of Z/n for some n. Groups with such a decomposition are called solvable.
Insolvability of the quintic then boils down to this: the symmetry group of any degree k polynomial must be a subgroup of S_k. For k <= 4, S_k is solvable; for k >= 5, it is not.
This last sentence is the closest thing I can offer to intuition. It is a very finite and tangible fact that one can write down and computationally verify for themselves. Like, you can write down the corresponding solvable decomposition of S_4, and you can see that S_5 is not solvable by verifying that A_5 is simple which itself boils down to a small number of cases.