Using group theory to prove Fermat's little theorem

Here is a demonstration of the use of group theory that I thought was quite neat. Fermat's little theorem states that if p is a prime number and a is a natural number that is not divisible by p (i.e. a is not congruent to 0 (mod p)) then
$$a^{p-1}\equiv 1 \;\mbox{(mod p)   ...(1) }$$
Now recall from M208 that if p is prime then
$$(Z^{*}_{p},\times_{p})$$
is a group of order p-1 (see the Handbook for M208 on page 30). Now we can use Lagrange's Theorem (Handbook page 35) to prove Fermat's little theorem as follows. Let a be a natural number that is not divisible by a prime number p, then a will be congruent to an element of the above group, i.e. to one of the set
$$\left\{1,2,3,..,(p-1)\right\}$$
Let this element be g so that
$$a\equiv g \;\mbox{(mod p)}$$
Now g will generate a cyclic subgroup of order n (Handbook page 28) and by Lagrange's Theorem (page 35) n will divide p-1, the order of the group. Hence
$$p-1=mn$$
for some natural number m and thus
$$a^{p-1}\equiv g^{p-1}\equiv g^{mn}\equiv(g^{n})^{m}\;\mbox{(mod p)}$$
Now by definition
$$g^{n}\equiv 1 \;\mbox{(mod p)}$$
and so
$$a^{p-1}\equiv 1^{m}\equiv 1\;\mbox{(mod p)}$$
Hence (1) is proven.