Considering $\mathbb Q_p$ the p-adic rationals; Tamely totally ramified extensions are obtained by adjoining solutions of the equation $x^e-pu=0$, where $e$ is the index of ramification and $u\in \mathbb Z_p^\times$.
Why is it true? I can't seem to understand how the roots of $x^e-pu=0$ are related to the demand $p\nmid e$ for tameness of ramification
If $K/\Bbb{Q}_p$ is a finite extension then its ring of integers $O_K$ is a DVR with uniformizer $\pi_K$ so that $\pi_K^e = pu$ for some $u \in O_K^\times$.
That $K/\Bbb{Q}_p$ is totally ramified means $O_K/(\pi_K) \cong \Bbb{F}_p$ so that $O_K = \{ \sum_{n=0}^\infty b_n \pi_K^n, b_n \in \{ \zeta_{p-1}^a\} \cup \{0\}\}= \sum_{n=0}^{e-1} \pi_K^n \Bbb{Z}_p$ and $[K:\Bbb{Q}_p]=[O_K:\Bbb{Z}_p]=e$. Note $u \in \zeta_{p-1}^a +\pi_K O_K$ for some integer $a$.
That $K/\Bbb{Q}_p$ is tamely ramified means $p \nmid e$.
This is important because for $p \nmid e$ then $(1+t)^{1/e} = \sum_{k=0}^\infty {1/e \choose k} t^k$ converges for $t \in \pi_K O_K$ so that $w=(u^{-1} \zeta_{p-1}^a)^{1/e} \in O_K^\times$ and hence $\varpi_K=w\pi_K= (p \zeta_{p-1}^a)^{1/e}$ is also an uniformizer of $O_K$ ie. $K = \Bbb{Q}_p(\varpi_K)$.
Finally $\varpi_K$ is a root of $x^e - \zeta_{p-1}^a p\in \Bbb{Q}_p[x]$ a polynomial of degree $[\Bbb{Q}_p(\varpi_K):\Bbb{Q}_p]$ which must be its minimal polynomial.