Hi Tina, for a lot of these proofs looking at some examples are probably best. If you can understand a few examples, then you probably understand the full idea.

Let's look at p=7 and a = 2. Try to do a few other examples, including other primes, such as p = 11, yourself.

Following the hint in the problem we want to look at the set $$2, 2\cdot 2, 3\cdot 2, 4\cdot 2, 5\cdot 2, 6\cdot 2.$$ and the product of all the elements in this set. Note that pulling out a 2 in each term, the product is just equal to $2^6\cdot 6!$. Simplifying each term in the set (mod 7), however, gives $$2, 4, 6, 1, 3, 5.$$ Note this is all the numbers 1-6, so the product of this is just 6!. Thus we have showed that $$2^6 \cdot 6! \equiv 6! \pmod{7}$$ Therefore, cancelling the 6! (which is possible as gcd(6!, 7) =1 so we can multiply by the mod. mult. inverse) we get $2^6 \equiv 1\pmod{7}$ which is what Fermat's Little Theorem says for $p = 7$ and $a = 2$.

As mentioned above, this is the main idea for any p and any a, so try to write out a few more examples to try to understand the general pattern.