Artin's Algebra (2nd edition): Proposition 2.4.3 I am self-studying algebra using MIT's publicly online materials.  The text for the course is the second edition of Artin's Algebra, Proposition 2.4.3 of which is stated without proof:

Let $x$ be an element of finite order $n$ in a group, and let $k$ be an integer that is written as $k = nq + r$ where $q$ and $r$ are integers and $0 \leq r < n$.  Then
  
  
*
  
*$x^k = x^r$.
  
*$x^k = 1$ if and only if $r = 0$.
  
*The order of $x^k$ is $\frac{n}{d}$, where $d$ is the greatest common divisor of $k$ and $n$.

1 and 2 are very easy, and I'm willing to bet that 3 is easy as well -- but I'm nonetheless stuck.  It's obvious that $(x^k)^{\frac{n}{d}} = 1$; I run into difficulty, however, trying to show that there is no positive integer $c < \frac{n}{d}$ such that $x^{kc} = 1$.
We certainly know that if there is such a $c$ then it must divide $\frac{n}{d}$.   Moreover, we know that $n = n'd$, $k = k'd$ for positive integers $n', k'$.  What I'm not seeing how to use, evidently, is the fact that $d = \gcd(n,k)$.  I'm sure the answer is straightforward, but I'm afraid I'm missing it.
Any help would be appreciated...
 A: If $c$ is the order of $x^k$, then $(x^k)^c=1$, that is $x^{kc}=1$. But the order of $x$ is $n$ and therefore $n\mid kc$, that is $n'd\mid k'dc$, and this is equivalent to $n'\mid k'c$. Since $(n',k')=1$, it follows that $n'\mid c$, that is $\frac nd\mid c$. On the other hand, $(x^k)^{\frac nd}=x^{\frac kdn}=1$. But $c$ is the order of $x^k$. So, $\frac nd=c$.
A: The order of $x^k$ is the smallest positive integer $r$ such that $x^{kr}=1$, i.e.  the smallest positive integer such that $kr$ is a multiple of $n$. 
As $kr$ is also a multiple of $k$, this means $kr$ is the least common multiple of $k$ and $n$. Now we have
$$kr=\operatorname{lcm}(k,n)=\frac{kn}{d}=k\,\frac nd,\quad\text{whence}\quad r=\frac nd.$$
A: Everything takes place in the cyclic subgroup generated by $x$, so it's not restrictive to assume $G=\langle x\rangle$, a cyclic group of order $n$.
Consider the homomorphism $f\colon\mathbb{Z}\to G$, $f(z)=x^z$, which is surjective and has kernel $n\mathbb{Z}$.
Now consider $g\colon\mathbb{Z}\to\mathbb{Z}$ defined by $g(z)=kz$. Since $f\circ g(1)=f(k)=x^k$, we have that the image of $f\circ g$ is $\langle x^k\rangle$ which we need to find the order of (the order of an element is precisely the number of elements in the cyclic subgroup it generates).
By general theory, the order is $l$, where $l\mathbb{Z}=\ker(f\circ g)$.
Now
$$
l\mathbb{Z}=\ker(f\circ g)=\{z\in\mathbb{Z}:g(z)\in\ker f\}=
\{z\in\mathbb{Z}:kz\in n\mathbb{Z}\}
$$
If $d=\gcd(n,k)$, set $n'=n/d$ and $k'=k/d$. Then
$$
kn'=k'dn'=k'n\in\mathbb{Z}
$$
so $n'\in l\mathbb{Z}$ and $l\mid n'$. Conversely, $kl=nt$ for some $t$, so we have $k'dl=n'dt$, and $k'l=n't$ so $n'\mid k'l$. Since $k'$ and $n'$ are coprime, we see that $n'\mid l$.
Therefore $l=n'$ as we wished to show.
A: Remember.  If $d=\gcd(k,n)$ then $\frac kd$ and $\frac nd$ are relatively prime.
Now together prop 1) and 2) prove that if $x^j = 1$ then $j$ is a multiple of $n$.
So if $(x^k)^m = x^{km} = 1$ then $km$ is a multiple of $n$ and $\frac kdm$ is a multiple of $\frac nd$.  And as $\frac kd$ and $\frac nd$ are relatively prime $m$ is a multiple of $\frac nd$.
So that's that.
If $m < \frac nd$ then $m$ is not a multiple of $\frac nd$ and $(x^k)^m \ne 1$.
So order of $x^k \ge \frac nd$. And as $(x^k)^{\frac nd} = (x^{n})^{\frac kd} = 1$ we have order of $x^k \le \frac nd$.
So order of $x^k$ is $\frac nd$.
A: Another option would be to use Proposition 2.3.5 c): There are some integers $r$ and $s$ with $d = rn+sk$. Suppose $x^k$ has order $m$. Then $x^{md}=x^{m(rn+sk)}=(x^n)^{mr}(x^k)^{ms}=1$ shows that $n \mid md$, so $\frac nd \mid m$. With $(x^k)^{\frac nd} = 1$ you have $m = \frac nd$.
A: $(x^{\large k})^{\large j}\!=1\!\!\! \overset{\large \ \,\color{#0a0}{(1)}}\iff\! n\mid kj \iff n\mid kj,nj\!\!  \overset{\large \ \,\color{#c00}{(2)}}\iff n\mid (kj,nj)\!=\!(k,n)j\iff   n/(k,n)\mid j$
$\color{#0a0}{(1)}\ $ holds by $\, n = {\rm ord}\, x\ $ so $\ x^i = 1\iff n\mid i$
$\color{#c00}{(2)}\ $ uses the GCD Universal Property $\ n\mid a,b\iff n\mid(a,b)\,$ and $ $ GCD Distributive Law
