Proving that ${1^n,2^n,3^n,…,(p−1)^n}$ is a reduced residue system modulo $p$ if $(n, p-1)=1$ My problem is identical to that of this person's. However, I'm not quite sure I understand the hint given in the best answer. Is there perhaps another method of solving this problem? I'm not very good with proofs, so any help is appreciated!
 A: Let $r$ be a primitive root of $p$, then the positive integers less than $p$ are congruent in some order to $r^1,r^2,\ldots ,r^{p-1}$ and the given set of integers will be congruent in some order to
$$r^{1n}, r^{2n},\ldots, r^{(p-1)n}$$
It is sufficient to show that none of these integers are congruent to each other in modulo $p$:
Suppose that  $r^{in}\equiv r^{jn} \pmod{p}$, where $1\le j\lt i\le p-1$. Because $r$ is a primitive root, $(r^{jn}, p)=1$ and we may divide $r^{jn}$ both sides and obtain $$r^{in-jn}\equiv 1\pmod{p}$$
But this holds only if the order of $r$ divides $in-jn$ : 
$$(p-1)|(in-jn)$$
Since we're given that $(p-1, n)=1$, we get
$$(p-1)|(i-j)$$ which is impossible unless $i=j$. 
That means the given set of integers are incongruent to each other and since we have $p-1$ incongruent integers modulo $p$, these must be congruent to $1,2,\ldots, (p-1)$ in some order. $\blacksquare$
A: Assuming that $n$ is relatively prime to $p-1$, then you know that there exist $a$ and $b$ such that $an+b(p-1)=1$. This let's you see that raising to the $n$th power is one-to-one. That is, it's not possible for $k_1^n\equiv k_2^n$ unless $k_1\equiv k_2$.
How exactly? Because the process is invertible by raising to the $a$: $(k^n)^a=k^{1-b(p-1)}=k\cdot(k^{p-1})^{-b}\equiv k\cdot1^{-b}=k$
So if raising to the $n$th power is one-to-one, and you start with all the $p-1$ residues that are relatively prime to $p$, then after raising to the $n$th power, you still have all the all the $p-1$ residues that are relatively prime to $p$.
A: If $n$ is arbitrary,
this is false.
For example,
if $n = 2$.
