$\newcommand{\p}{\mathfrak{p}}
\newcommand{\O}{\mathcal{O}}$
Let $K$ be any number field. Any prime ideal $\p$ of $\O_K$ determines a discrete valuation $v_\p$ on $K$: for $x\in \O_K$, define $v_\p(x) = n$, where $n$ is the highest integer such that $x\in \p^n$, where $\p^0=\O_K$. For arbitrary $\alpha\in K$, write $\alpha=\frac xy$ for $x,y\in \O_K$, define $v_\p(\alpha)=v_\p(x)-v_\p(y)$. You can normalise this differently: pick any $k\in \mathbb{N}$, define $v_\p(x) = n/k$, where $n$ is as above.
If you set $k=1$, then the valuation is normalised in such a way as to have image $\mathbb{Z}$. But in general, its restriction to $\mathbb{Q}$ will not have image $\mathbb{Z}$, so such a normalisation will not extend a normalised valuation from $\mathbb{Q}$. If you want $v_\p$ to extend $v_p$, where $p$ is the rational prime lying below $\p$, then you need to take $k=e_\p$, the ramification index of $\p/(p)$. In this case, of course, the image of $v_\p$ is not $\mathbb{Z}$, but $\frac 1e\mathbb{Z}$.
In the example you give, $p$ is totally ramified, so there is exactly one prime in $\O_K$ lying above $p$, generated by $\zeta_p-1$ (to convince yourself of that, compute the norm of this elements as the constant coefficient in its minimal polynomial), and the extension of $v_p$ to $K$ is unique. In general, there are as many non-equivalent extensions of $v_p$ to $K$, as there are prime ideals of $\O_K$ lying above $p$.
Best Answer
In order for this question to stop being considered unanswered, I will copy the above comments by KCd:
"Squares in $\mathbb Z_p$ are simpler than in $\mathbb Z$, not more complicated! Anyway, just knowing the ramification index and residue field degree in general is insufficient. A general answer is in Lang's Algebraic Number Theory (2nd ed.), Prop. 23 on p. 26. Using any uniformizer $\pi$ (which you could check if you know the ramification index) and generator $\gamma$ of the residue field extension (which you could check if you know the residue field degree), the integers of the extension are $\mathbb Z_p[\pi,\gamma]$. If $e=1$ use $\pi=p$, so the ring is $\mathbb Z_p[\gamma]$. If $f=1$ use $\gamma=1$, so the ring is $\mathbb Z_p[\pi]$. As for the special case of $\mathbb Q_p(i)$, the ring of integers is in general $\mathbb Z_p[i]$, even for $p=2$, but notice $\mathbb Z_p[i]=Z_p$ if $p\equiv 1\pmod4$ since $−1$ is a $p$-adic square in that case, so writing the ring of integers as $\mathbb Z_p[i]$ is kind of misleading as to what the ring looks like.
Another thing to be careful about is that you know the degree of your field over $\mathbb Q_p$! For instance, asking about the ring of integers of $\mathbb Q_5(\sqrt[3]{2})$ is ambiguous, because $X^3−2$ has one root in $\mathbb Q_5$ and the other roots are quadratic over $\mathbb Q_5$."
Of course, if KCd wants to reclaim his answer (as I suggested above in the comments), I will gladly delete this CW answer.