Finding $p^\textrm{th}$ roots in $\mathbb{Q}_p$? So assume we are given some $a\in\mathbb{Z}_p^\times$ and we want to figure out if $X^p-a$ has a root in $\mathbb{Q}_p$. We know that such a root must be unique, because given two such roots $\alpha,\beta$, the quotient $\alpha/\beta$ would need to be a non-trivial $p^\textrm{th}$ root of unity and $\mathbb{Q}_p$ does not contain any.
Now we can't apply Hensel, which is the canonical thing to do when looking for roots in $\mathbb{Q}_p$. What other approaches are available? 

Let's take $\alpha\equiv 1\mod p^2$. Then $\alpha = 1 + p^2\beta\in\mathbb{Z}_p$, where $\beta\in\mathbb{Z}_p$. What we want to do is make Hensel's lemma work for us, after changing the equation a bit. We have $f(x) = x^p - \alpha = x^p - (1 + p^2\beta)$. We see that if $x$ is to be a solution, it must satisfy $x\equiv 1\mod p$, so $x = 1 + py$ for some $y\in\mathbb{Z}_p$. Now we have a new polynomial equation:
$$
f(y) = (1 + py)^p - (1 + p^2\beta) = \sum_{i = 0}^p \begin{pmatrix} p \\ i\end{pmatrix}(py)^i - (1 + p^2\beta),
$$
which reduces to
$$
f(y) = \sum_{i = 1}^p \begin{pmatrix} p \\ i\end{pmatrix}(py)^i - p^2\beta.
$$
So long as $p\neq 2$, we can set this equal to zero and cancel a $p^2$ from each term, and get
$$
0 = p^2 y + \begin{pmatrix} p \\ 2\end{pmatrix}(py)^2 + \ldots + (py)^p - p^2\beta = y + \begin{pmatrix} p \\ 2\end{pmatrix} y^2 + \ldots + p^{p-2}y^p - \beta.
$$
Mod $p$, we can solve this equation:
$$
y + \begin{pmatrix} p \\ 2\end{pmatrix} y^2 + \ldots + p^{p-2}y^p - \beta \equiv y - \beta \equiv y - \beta_0\mod p,
$$
where $\beta = \beta_0 + \beta_1 p + \beta_2 p^2 + \ldots$ by $y = \beta_0$. Mod $p$, our derivative is always nonzero:
$$
\frac{d}{dy}\left[y - \beta_0\right] \equiv 1\mod p,
$$
so we can use Hensel's lemma and lift our modified solution mod $p$ to a solution in $\mathbb{Q}_p$. Therefore, if $\alpha\in 1 + p^2\mathbb{Z}_p$ and $p\neq 2$, there exists a $p$th root of $\alpha$ in $\mathbb{Q}_p$.
