There is a primitive $m^{th}$ root of unity in $\mathbb{Q}_p$ $\Leftrightarrow m \mid (p-1)$ After Hensel's Lemma there is the following proposition in my notes: 

If $p$ is a prime and $m \in \mathbb{N}$ then there is a primitive $m^{th}$ root of unity in $\mathbb{Q}_p$ $\Leftrightarrow m \mid (p-1)$. 

To prove this proposition we begin as follows: 
$(\Rightarrow) $ Let $a$ be a primitive $m^{th}$ root of unity in $\mathbb{Q}_p$. That means that $m$ is the smallest natural number such that $$a^m \equiv 1 \pmod p$$ (correct?) 
Can we suppose that $m$ and $p$ are coprime, and use Fermat's little theorem we have that $a^{p-1} \equiv 1 \pmod p$? 
Can we conclude from this that $m \mid (p-1)$? 
How do we prove the other direction? Perhaps using Hensel's Lemma?
 A: The easiest way to prove this is as follows:
You want to ask if the polynomial
$$x^m-1$$
has a root in $\Bbb Q_p$. We know if you pick one such primitive root, $\zeta$ then we can get all of the others by examining $1,\zeta,\zeta^2,\ldots, \zeta^{m-1}$.
Now, reducing this polynomial modulo $p$ we see that we want to know if there are $m^{th}$ roots of $1$ in $\Bbb F_p$, however we know by the classification of finite fields that the elements of $\Bbb F_p$ are exactly those that satisfy $x^p-x=0$. Now, since roots of unity are not $0$, we see all non-zero elements of $\Bbb F_p$ satisfy $x^{p-1}-1$. More completely we have the result from the structure of finite fields

If $F$ is a finite field of order $q$, then $F^\times$ is a cyclic group of order $q-1$.

From this we see that for each divisor $d|(q-1)$, we may take a multiplicative generator of $F^\times$, say we call it $g$, and note that $g^{(q-1)/d}$ is an element of order $d$ in the group, i.e. it is a $d^{th}$ root of $1$. And if $\zeta\in F$ is an $m^{th}$ root of one, then the multiplicative group $\langle \zeta\rangle\le F^\times$ must have order dividing the order of the group, i.e. $m|(q-1)$. So it is necessary and sufficient that $m|(q-1)$ for a finite field, $F$, of order $q$ to possess the $m^{th}$ roots of $1$.
Now, in our case the residue field is just $\Bbb F_p$, so we see it has all $m^{th}$ roots of $1$ for $m|(p-1)$ and no others. This implies immediately that if $m\not\mid (q-1)$, then $\Bbb Q_p$ does not possess $m^{th}$ roots of $1$, since if we had them in the bigger field, quotienting to the residue field would still keep them.
To see the other direction, we note that if $m|(p-1)$, in particular $m$ is coprime to $p$, hence since
$${d\over dx}(x^m-1)=mx^{m-1}$$
and $|m|_p=1$ so we satisfy the hypotheses of Hensel's lemma, and get a lifting back up to $\Bbb Q_p$. This completes the proof.
