Factoring $t^{q^r}-t$ over the finite field $\mathbb{F}_q$ This might be a very short-lived question.
Let $F$ be a finite field of order $q$ ($q$ is a prime power). It is well known that the polynomial$$t^{q^d}-t$$factors into the product of all irreducible polynomials over $F$ of degree $r$ where $r$ divides $d$.
My question is why is this? I know this is basic, but still, whenever I try to search for this on the internet, I get the answers about factoring arbitrary polynomials where this is only mentioned as a known fact. I had no luck finding the answer to my question.
I do understand that if we consider an extension field $E/F$ with $q^d$ elements then this polynomial factors into linear factors. I noticed that $t^{q^r}-1$ divides $t^{q^d}-1$ iff $r$ divides $d$. What ideas do I miss to understand the statement I am asking about?
 A: The general version of your observation towards the end is that $t^\ell - 1 \mid t^m - 1$ if and only if $\ell \mid m$; this is useful (and follows essentially from the division algorithm). Note that this divisibility is true if we evaluate the polynomials too.
With this in hand: Let $f(t) \in F[t]$ be an irreducible polynomial of degree $r$.
Then $r \mid d$ if and only if $f(t) \mid t^{q^d} - t$.
Proof. Forwards, assume $r \mid d$. Then $q^r - 1 \mid q^d - 1$, so $t^{q^r - 1} - 1 \mid t^{q^d - 1} - 1$.
Multiply through by $t$ and we have $t^{q^r} - t \mid t^{q^d} - t$.
Now since $f(t) \in F[t]$ is irreducible of degree $r$, if we let $\alpha \in \bar{F}$ be a root of $f(t)$, we have $[F(\alpha) : F] = r$, making $| F(\alpha) | = q^r$, and $F(\alpha)$ is the splitting field of $t^{q^r} - t$ over $F$ by uniqueness of finite fields.
Hence $\alpha$ is also a root of $t^{q^r} - t$, so $f(t) \mid t^{q^r} - t \mid t^{q^d} - t$.
Conversely, assume $f(t) \mid t^{q^d} - t$ with $f(t)$ irreducible of degree $r$.
Again let $\alpha \in \bar F$ be a root of $f(t)$, so that $f(t) \mid t^{q^d} - t$ implies $F(\alpha) \subset E$ where $E$ is the splitting field of $t^{q^d} - t$ (so $|E| = q^d$).
Then
$$ d = [E : F] = [E : F(\alpha)] \cdot [F(\alpha) : F] = [E : F(\alpha)] \cdot r, $$
meaning that $r \mid d$.
