Help me please to prove that:
For any $k \in \mathbb{N}$ each element of field $F_q$ is the $k$-th power of some element from that field if and only if $GCD(q-1, k)=1$.
My approach
Let's look at multiplicative group of that field $F_q^*$. That group is cyclic, so it has $q-1$ elements and they all generated by some element $\alpha$. So, $F_q^*$ looks like $\{\alpha, \alpha^2, \dots, \alpha^{q-1}\}$. If we raised them all to the power $k$, we'll get $\{\alpha^{k}, \alpha^{2k}, \dots, \alpha^{(q-1)k}\}$. Now we need to prove that all that raised elements are different if and only if $GCD(q-1,k)=1$. But I don't know how to do this.
Let $x$ be a non-zero element of $F_q$. If $k$ and $q-1$ are relatively prime, there are integers $s$ and $t$ such that $ks+(q-1)t=1$ (Bezout). So $$x^1=(x^s)^k(x^{q-1})^t=(x^s)^k.$$
For the other direction, show that if the gcd of $k$ and $q-1$ is not $1$, then the map that takes $t$ to $t^k$ is not one to one.