Let $\mathbb{Q}_p(\zeta_p)$ be the cyclotomic extension of the $p$-adic field $\mathbb{Q}_p$.
Then $\pi=\zeta_p-1$ is an uniformizer of the ring of integer $\mathbb{Z}_p[\zeta_p]$ as $v_p(\zeta_p-1)=\frac{1}{p-1}$.
We also have $\pi=\sqrt[(p-1)]{-p}$ is also an uniformizer which has valuation $\frac{1}{p-1}$.
In fact, any two uniformizer are conjugate to each other. That means, if $pi$ and $\pi'$ be two uniformizer then there is a unit $u \in \mathbb{Z}_p[\zeta_p]^{*}$ such that $$\pi'=u \pi.$$ Thus if $\pi=\sqrt[(p-1)]{-p}$ is an uniformizer, then $\pi=\sqrt[(p-1)]{p}$ should be an uniformizer.
Am I right saying that $\pi=\sqrt[(p-1)]{p}$ is also an uniformizer of $\mathbb{Z}_p[\zeta_p]$ ?
Now suppose $p=2$,
Is at least $\sqrt{-1} \in \mathbb{Q}_2(\zeta_2)$ so that $\pi=\sqrt{2}$ will be an uniformizer of $\mathbb{Z}_2[\zeta_2]$ ?
Is there any modification of $\pi=\sqrt[(p-1)]{-p}$, so that it becomes uniformizer of $\mathbb{Z}_2[\zeta_2]$ ?
This will work as long as $\sqrt[p-1]{-1}\in K=\Bbb Q_p(\zeta_p)$. The trouble is, it isn't. The roots of unity of order coprime to $p$ in $K$ are the same as those in $\Bbb Q_p$ and all satisfy $X^{p-1}=1$ (as long as $p$ is odd). So there are no solutions of $X^{p-1}=-1$ in $K$.