This result (Abel's theorem) is valid for any field of characteristic $0$, and relies on the following result, proved by Galois:
A polynomial equation over a field of characteristic $0$ is solvable by radicals if and only if its Galois group is solvable.
Now with literal coefficients, the Galois group of a polynomial of degree $n$ is the symmetric group $S_n$, which is not solvable for $n\ge 5$. For details on what is a
solvable group, you can see Wikipedia.
The situation is more complex in characteristic $p>0$, because if a root has order divisible to the characteristic, it gives rise to a so-called non-separable extension, i.e. it is not a simple root of its minimal polynomial, contrary to the case of characteristic $0$.