THEOREM 18.18 Let K be a field of characteristic 0 and let L : K be a finite normal extension with soluble Galois group G. Then there exists an extension R of L such that R : K is radical.
I'm studying this proof using the "Galois Theory" by Ian Stewart and I have doubts about some statements. The proof is done by induction on $n = |G|$. According the text, the result is clear when $n = 1$. Why? If $n = 1$, then we know that $[L : K] = 1$ and $L = K$. Plus $G = \{id\}$. But what does make the result clear?
Soon after, the text sets a new field $N$ which is the splitting field of the polynomial $t^p - 1$ over $L$ and once $L : K$ is normal, then $L$ is the splitting field of some polynomial $f$ over $K$. So far, its all good. But then, the text states that $N$ is also the splitting field of $(t^p - 1)f$ over $L$. Why is this last statement true?
I really appreciate all kind of help, because I'm really stuck.
Thank you!