Let $K$ be a nonarchimedean local field, $L$ be the unique degree $n$ unramified extension of $K$. Let $\sigma$ be the Frobenius element of $L/K$, that is, $\sigma\in\mathrm{Gal}(L/K)$ such that $\sigma(x)\equiv x^q(\text{mod }\mathfrak{m}_L)$ for every $x\in \mathcal{O}_L$, where $\mathcal{O}_L$ is the valuation ring of $L$, $\mathfrak{m}_L$ is the unique maximal ideal of $\mathcal{O}_L$, and $q$ is the size of residue field of $K$.
My question is: if $\alpha\in \mathcal{O}_L^\times$ has $\mathrm{Nm}_{L/K}(\alpha) = 1$, must there be a $u\in \mathcal{O}_L^\times$ such that $\dfrac{\sigma(u)}{u} = \alpha$?
$\mathrm{Nm}_{L/K}(\alpha) = 1$ means that $\alpha\sigma(\alpha)\cdots\sigma^{n-1}(\alpha) = 1$, so $\alpha^{1+q+\cdot+q^{n-1}}\equiv 1(\text{mod }\mathfrak{m}_L)$, that is, the order of $\alpha\text{ mod }\mathfrak{m}_L$ divides $1+q+\cdots+q^{n-1}$. As the multiplicative group of the residue field of $L$ is cyclic of order $q^n-1$, we se that $\alpha\text{ mod }\mathfrak{m}_L$ is a $(q-1)-$th power, so by Hensel's lemma $\alpha$ is a $(q-1)-$th power in $\mathcal{O}_K$. Write $\alpha = u_0^{q-1}$, then at least we get
$$\dfrac{\sigma(u_0)}{u_0}\equiv\alpha (\text{mod }\mathfrak{m}_L).$$ But, suppose that I want to proceed to find $u_1\equiv u_0 (\text{mod }\mathfrak{m}_L)$ such that $\dfrac{\sigma(u_1)}{u_1}\equiv\alpha (\text{mod }\mathfrak{m}^2_L)$, I choose a prime element $\pi$ of $\mathcal{O}_K$, write $u_1 = u_0 + s\pi$, then we need
$$s^q - \alpha s + \dfrac{\sigma(u_0)-\alpha u_0}{\pi} \equiv 0 (\text{mod }\mathfrak{m}_L),$$
How do we know if this degree $q$ equation is solvable over $\mathcal{O}_L/\mathfrak{m}_L$? And, if this is solvable, can we proceed like this to find $u_2, u_3, \cdots$?