Suppose that $n$ is a natural number and p is a prime that does not divide n. Let $L$ be the splitting field of the polynomial $X^n-1$ over $\mathbb{F}_p$. Show that $[L:\mathbb{F}_p]$ is the smallest natural number k s.t. $p^k=1$ mod n.
Solution: We know that $\#L=p^k$ for some k and $L$ is absolutely determined by that. We also know that the multiplication group $L^*$ is cyclic $p^k-1$ group. So $L$ has n solution to the equation $x^n=1$ iff n divides $p^k-1$. Thus $p^k=1$ mod n.
First I have not been able to prove the step that $x^n=1$ iff n divides $p^k-1$. I thought of looking at it as integers with respect to sum module $p^k-1$. Then the equation is of the form $nx=0$ and then we want to show that this has n solutions iff n divides $p^k-1$.
Secondly, in this solution I find no attempt to show that k is the smallest number but I guess that is the property of the splitting field, if it were not the smallest number then it would be a splitting field for other irreducible polynomial over $\mathbf{F}_p$