Computing a uniformizer in a totally ramified extension of $\mathbb{Q}_p$.

Do you know how to compute a uniformizer of $\mathbb{Q}_p(\zeta_{p^n},p^\frac{1}{p})$?

Where $\zeta_{p^n}$ is a primitive $p^n$-th root of 1 and $p$ is an odd prime.

A uniformizer of $\Bbb Q_p(\zeta_{p^n})$ is $x = \zeta_{p^n}-1$, as its minimal polynomial is of the form $x^{\phi(p^n)} + pxP(x) + p$ for some polynomial $P \in \Bbb Z[x]$.

In particular, $p = -x^{\phi(p^n)} + O(px)$

Assuming $n \ge 2$, we have $p = (-x^{\phi(p^{n-1})})^p + O(px)$, so it's more useful to view $\Bbb Q_p(\zeta_{p^n},p^\frac 1p) = \Bbb Q_p(\zeta_{p^n},u = (-p/x^{\phi(p^n)})^\frac 1p)$

We can further try to divide $-p/x^{\phi(p^n)}$ by a $p$-th power $k^p$ to make $u' = u/k$ closer to $1$. Unfortunately I don't know enough about the $x$-adic development of $-p$ to have a guess that would work all the time.

Then taking $v = u'-1$ we obtain $v^p = O(pv) + (u'-1)$, where hopefully the $x$-valuation of $(u'-1)$ is not a multiple of $p$ and is still less than that of $pv$, which would make a combination of powers of $v$ and $x$ into a uniformizer.

Here are the details with $\Bbb Q_3(\zeta_9,3^{1/3})$.

Let $x = \zeta_9 - 1$ and $u = -3^{1/3}/x^2$. We have $u^3 = -3/x^6 = 1 + 2x^3 + x^4 + O(x^5)$.
We pick $k = -2-x$ so that $u'^3 = -3/k^3x^6 = 1 + x^4 + O(x^5)$.
We let $v = u'-1$, so $v^3 = x^4 + O(x^5)$.

Then $|v|=4|x|/3$ so $w = v/x$ is a uniformizer.

To get a reality check, $w = (3^{1/3}-2x^2-x^3)/(2x^3+x^4)$, and its minimal polynomial is $w^{18} + 9w^{17} + 45w^{16} + 180w^{15} + 648w^{14} + 1782w^{13} + 3651w^{12} + 5814w^{11} + 7569w^{10} + 8370w^9 + 7956w^8 + 6471w^7 + 4473w^6 + 2592w^5 + 1233w^4 + 468w^3 + 135w^2 + 27w + 3 $

hence $3 = -w^{18} + O(3w)$ as we wanted.

(for some reason picking $k=1+2x$ doesn't give an algebraic integer in the end so I couldn't do the check)

I gave this question to a student as a summer project. Thanks to Mercio's answer, we have worked out how to find explicit uniformizers of $\mathbb{Q}_p(\zeta_{p^2},p^{\frac{1}{p}})$ for all odd $p$. We have written a short article on our calculations, which can be found here: http://arxiv.org/abs/1912.01656 Please let us know if you have any suggestions and/or comments for us.