field whose every non-zero element is a root of unity In his notes Algebraic Number Theory 2 by Pete Clark, the author asks following exercise:

Show that for a field $k$, TFAE:


(1) Every nonzero element of $k$ is a root of unity.


(2) The characteristic of $k$ is  $p> 0$, and $k/\mathbb{F}_p$ is algebraic.

I couldn't proceed for the solution. Suppose we want to show $(1)\Rightarrow (2)$. If we go by norms on fields, then, since every non-zero element of $k$ is a root of unity, so its norm must be $1$. Thus, the field $k$ admits only trivial norm:
$|x|=1$ if $x\neq 0$ and $|x|=0$ if $x=0$.
This did not give any contradiction to me, since every field - whether of characteristic $0$ or positive - always admits the trivial norm defined in above way.
I couldn't proceed for proof. Any hint (and hint only will be sufficient).
 A: Notation: $k^\times := k \setminus \{0\}$.

(1) $\Rightarrow$ (2): If $k$ has characteristic $0$, then $a = 1 + 1 \in k$ is not a root of unity. (Why?)
Now, assuming that the characteristic is $p$, given any $\alpha \in k^\times$, $\alpha$ satisfies the polynomial $X^n - 1 \in \Bbb F_p[x]$ for an appropriate $n$ (which $n$?).
(Note that $\Bbb F_p$ is a subfield of $k$ in a natural way.)

(2) $\Rightarrow$ (1): Let $\alpha \in k^\times$. Then, $F = \Bbb F_p(\alpha)$ is a finite extension of $\Bbb F_p$. (Why?)
Let $n := |F|$. (This is finite. Why?)
Then, $F^\times$ is a group of order $n - 1$ containing $\alpha$.
Thus, $\alpha^{n - 1} = 1$.
A: If the characteristic is zero, then there is an element that's not a root of unity, because $2^n\ne1$, for every $n$, in the rational numbers.
Thus the statement (1)$\Rightarrow$(2) is obvious, because if every element is a root of unity, it is algebraic over the prime subfield.
Suppose $k/\mathbb{F}_p$ is an algebraic extension and take $a\in k$, $a\ne0$. Then $\mathbb{F}_p(a)$ is a finite field and if $q$ is its cardinality, every element therein is a root of the polynomial $x^q-x$. Since $a\ne0$, it is a root of $x^{q-1}-1$
