$x^5-1$ completely splits in $\mathbb F_{16}$

Question

I need to prove that $x^5-1$ completely splits in $\mathbb F_{16}$. This means it has exactly $5$ unique roots in $\mathbb F_{16}$. I have only found the following way: find an irreducible polynomial of degree $4$ over $\mathbb F_2[x]$, write all $16$ remainders of it, raise all to power $5$, and see that for exactly $5$ of them I got $1$. I believe there are other (short) proofs. Can you please give me a hint?

Answer 1

Hint: the multiplicative group $\mathbb{F}_{16}\setminus\{0\}$ is cyclic and has order $15$.

Answer 2

Let $g(x) = (x^5-1)/(x-1)$. Show that $g(x)$ is irreducible in $\mathbb F_2[x]$. Show that $\mathbb F_2[x] / (g(x))$ is the splitting field of $g(x)$ (this is true for finite fields in general). What's the order of $\mathbb F_2[x] / (g(x))$?

Answer 3

The good (+1) answer by egreg is the easiest route to the destination I think. An alternative is to prove that $$ x^4+x^3+x^2+x+1=\frac{x^5-1}{x-1} $$ is irreducible in $\Bbb{F}_2[x]$. Therefore we can construct the field of 16 elements as $$ \Bbb{F}_{16}=\Bbb{F}_2[x]/\langle x^4+x^3+x^2+x+1\rangle, $$ and the coset of $x$ is a primitive fifth root of unity there.