Having calculated the fifth roots of unity $1, w, w^2, w^3, w^4$, i.e. the solutions to the equation $ u^5 = 1$, we can recast our equation as $ \left( \frac{z+1}{z-1} \right) ^5 = 1$ provided $z$ is not equal to $1$.
It is easy to verify that $z = 1$ is not a solution to the equation $(z+1)^5 = (z-1)^5$. So for $z$ not equal to $1$, we have that $ \frac{z+1}{z-1} = 1, w, w^2, w^3, w^4 $. Simple manipulation will allow you to express $z$ in terms of these roots.
Note that setting $ \frac{z+1}{z-1} = 1 $ will not yield a solution. This can be expected, since the equation we are solving is quartic and so has at most 4 distinct roots.
I am going to take as an assumption that your statement, that when $a_m\neq 0$, all roots have multiplicity $ 1 $, is true. It certainly sounds correct, but I'm no expert on that part.
However, you can easily show the case for $a_m=0$ from that point, as follows:
Suppose that the smallest index that arises is $p^e$. That is,
$$
f(x) = \sum_{i=0}^{m-e} a_ix^{p^{m-i}}
$$
where $a_0=1$ and $a_e\neq 0$. Now, let $y=x^{p^e}$. This allows us to write
$$
f(x) = \sum_{i=0}^{m-e} a_i y^{p^{m-i}/p^e} = \sum_{i=0}^{m-e} a_i y^{p^{m-e-i}/p^e}
$$
Let $M=m-e$, so we have
$$
f(x) = \sum_{i=0}^M a_i y^{p^{M-i}} = g(y)
$$
Now, $g(y)$ has, from the assumed statement, $p^M$ roots of multiplicity $ 1 $.
Now is where we bring in the characteristic being $p$, which means that $(x+y)^p = x^p+y^p$. So we seek solutions to $x^p=a$. If $x_1^p=x_2^p=a$, then $x_1^p-x_2^p = (x_1-x_2)^p=0$, and so $x_1=x_2$. So $x^p-a=0$ has one root of multiplicity $p$, and by iterating this process, we find that $x^{p^e}-a=0$ has one root of multiplicity $p^e$.
But our roots of $g(y)=f(x)$ take the form $y=x^{p^e}=a$, so each root of $g(y)$ is associated uniquely with a root of $f(x)$, and has multiplicity $p^e$.
Best Answer
Consider the polynomial $x^n-a\in \mathbb Q[x]$, with $a>0$. One root of this polynomial is $\alpha$, the $n$-th root of $a$. Now let $\omega\in \mathbb C$ be a primitive $n$-th root of unity. That means that $\omega ^n=1$ and $\omega^k\ne 1$ for all $0<k<n$. So, the (typically complex) numbers $\beta_i=\omega ^i\alpha $, where $0\le i < n$, are distinct and all satisfy: $\beta _i^n=(\omega ^i\alpha )^n=(\omega ^n)^i\alpha ^n=a$. So, this shows that the polynomial has the $n$ distinct roots $\beta _0,\cdots ,\beta _n$.
(Partly) Conversely, if a polynomial $p(x)\in \mathbb Q[x]$ is of degree $n$, and $\alpha$ is a root of $p(x)$ such that for some primitive $n$-th root of unity $\omega $, it holds that $\omega^i\alpha$ is a root of $p(x)$ for all $i$, then $p(x)$ splits as $(x-\alpha)(x-\omega \alpha)(x-\omega ^2 \alpha)\cdots (x-\omega ^{n-1}\alpha)$ which is the splitting of the polynomial $x^n-(\alpha^n)$. Writing $a=\alpha ^n$ this shows that $p(x)=x^n-a$.