I am looking at this Theorem and proof from : https://www.researchgate.net/publication/320835842_The_Inverse_Galois_Problem_4th_year_project (page 12)
For n > 3, there exists a polynomial $f \in \mathbb{Q}[t]$ with splitting field $\mathbb{L}$ over $\mathbb{Q}$ such that Gal($\mathbb{L} : \mathbb{Q}$) = S$_n$.
Proof. By lemma 3.2 above, we may choose three polynomials $f_1; f_2$ and $f_3$ of degree n that satisfy the following conditions: $f_1$ is irreducible modulo 2; $f_2 \equiv g_1g_2$ (mod 3), where $g_1$ is irreducible modulo 3 with deg($g_1$) = n - 1, and deg($g_2$) = 1; $f_3 \equiv h_1h_2...h_l$ (mod 5), where $h_1; ... ; h_l$ are all irreducible modulo 5, deg($h_1$) = 2, and the number of $h_i$ with odd degree is either 1 or 2. Now we set $f = 15f_1 + 10f_2 + 6f_3$ so that $f \equiv f_1$ (mod 2); $f \equiv f_2$ (mod 3); and $f \equiv f_3$ (mod 5); and let $\mathbb{L} : \mathbb{Q}$ be the splitting field extension for $f$ over the rationals. Denote by G the Galois group Gal($\mathbb{L} : \mathbb{Q}$) of $f$. Then one finds that $f$ is irreducible modulo 2, and is therefore irreducible over $\mathbb{Q}$ (by Gauss' lemma), so it follows that G is isomorphic to a transitive subgroup of S$_n$. Moreover, G contains an (n - 1)-cycle, as $f \equiv g_1g_2$ (mod 3), there exists a $\sigma \in $ G such that $\sigma$ permutes the n-1 roots of $g_1$ in $\mathbb{L}$, and contains a transposition multiplied by odd cycles, as $f \equiv h_1...h_l$ (mod 5)... (it keeps going)
I am not sure I understand why G must contain an (n-1)-cycle. The proof makes it seem obvious.
A similar proof can also be found here: http://www.ru.ac.bd/stat/wp-content/uploads/sites/25/2019/03/104_09_01_van-der-Waerden-B.L.-Modern-Algebra-I-1949.pdf (page 191)
But this one also kind of states the fact with explaining why...
Any explanation would be greatly appreciated.
The Galois group $H$ of the splitting field of this polynomial over $\mathbb{F}_3$ is a subgroup of the Galois group $G$ over $\mathbb{Q}$; the proof of this is in Chapter 61, page 190 of van der Waerden's text which you link. However, any extension of finite fields is automatically Galois, and the Galois group is cyclic, generated by Frobenius. Therefore $H \subseteq G$ must be isomorphic to $\mathbb{Z}/(n - 1)\mathbb{Z}$.