Show that the equation has exactly $m$ different roots in the algebraic closure  Let $n=p^rm$, where $p$ is a prime, $m \in \mathbb{N}, r \geq 0$ an integer and $(p,m)=1$. 
 I have to show that the equation $x^n=1$ has exactly $m$ different roots in the algebraic closure $\overline{\mathbb{Z}}_p$ of $\mathbb{Z}_p$.   
I have done the following:   
In $\mathbb{Z}_p$ it stands that $x^p=x$.   
So, we have that  $$x^{p^r}=(x^{p})^{p^r-1}=x^{p^r-1}=(x^p)^{p^r-2}=x^{p^r-2}= \dots =x^p=x$$
 That means that $x^n=1 \Rightarrow x^{p^rm}=1 \Rightarrow (x^{p^r})^m=1 \Rightarrow x^m=1$ 
  Is this correct?? Do we conclude from that, that the equation $x^n=1$ has exactly $m$ different roots in the algebraic closure $\overline{\mathbb{Z}}_p$ of $\mathbb{Z}_p$ ??  
 A: Your calculations are correct. They show that every $n$-th root of $1$ is an $m$-th root of $1$. We can round out your argument as follows.
You showed that $x^n = 1$ implies $x^m = 1$. In the other direction, we have that $x^m=1$ implies $x^n=1$ since $m|n.$ So, in order to show that the polynomial $Y^n-1 \in \overline{\mathbb Z_p}[Y]$ has exactly $m$ different roots, we must show that the polynomial $Y^m-1 \in \overline{\mathbb Z_p}[Y]$ has exactly $m$ different roots. There's a "well known" criterion for that, namely a polynomial $f(Y) \in \overline{\mathbb Z_p}[Y]$ doesn't have multiple roots iff $gcd(f(Y),f'(Y)) = 1.$ Here, $f'$ is the (formal) derivative of $f.$ Also, recall that $\overline{\mathbb Z_p}[Y]$ is a Euclidean ring, so the gcd is well defined. We calculate
$$
\left(Y^m-1\right)' = mY^{m-1}.
$$
Since $p\not|\ m$ (you should state this in your question), $mY^{m-1} \neq 0$, and the only root is $0$. But $0$ is not a root of $Y^m-1$, so we have indeed $gcd(Y^m-1,\left(Y^m-1\right)') = 1$, and $Y^m-1$ has exactly $m$ different roots in $\overline{\mathbb Z_p}$, and $Y^n-1$ also  has exactly $m$ different roots in $\overline{\mathbb Z_p}$.
Let me give some more technical details. It is "well known" that a polynomial $f$ of degree $n$ over some field $K$ has at most $n$ roots (counted with multiplicity). If $K$ is algebraically closed, $f$ has excatly $n$ roots (counted with multiplicity). Moreover, if $K$ has characteristic $p > 0,$ we have in $K[Y]$
$$
Y^{mp^r}-1 = (Y^m-1)^{p^r}.
$$
Basically, you have proved this with the calculations in your question. From this identity, we see that the roots of $Y^{mp^r}-1$ are exactly the roots of $Y^m-1$, but each with multiplicity $p^r.$
