Why does the multiplicative order divide $|\mathbb{F}|-1$?

Question

This source describes Lagrange's theorem for finite fields (page 6). It generalizes Fermat's little theorem. In particular, it states that if $\mathbb{F}$ is a field with finitely many elements, with $|\mathbb{F}|=m$, then $$a^{m-1}=1$$ for every $a\in\mathbb{F}\setminus\{0\}$.

To prove this, they note that $\mathbb{F}\setminus\{0\}$ has $m-1$ elements. But then, they suddenly conclude that this implies $\text{ord}_{\mathbb{F}}(a)$ divides $m-1$. The result then follows.

How is the divisibility claim reached?

Answer 1

$\mathrm{ord}_{\Bbb F}(a)=|\langle a\rangle|\mid|\Bbb F\setminus\{0\}|$ for any $a\in\Bbb F\setminus\{0\}$. This is Lagrange's theorem for groups

Answer 2

In a field, the nonzero elements always form a group (under multiplication). The order of that group is of course $|\Bbb F|-1$. The conclusion is then reached by Lagrange's theorem.

Indeed Fermat's little theorem then follows by application to the field $\Bbb Z_p$, for any prime $p$.