I don't understand a step in the proof of Euler's Theorem, please explain I am trying to learn the proof for Euler's theorem which states:

If $\gcd(a,m)=1$ then $a^{\phi(m)} \equiv 1 \mod m$. 

The proof goes like this. Take the reduced residue system modulo $m$. $ar_1,...,ar_k$ is also a reduced residue system modulo $m$ (the text proves this lemma) then they state that, 
$$r_1 r_2...r_k \equiv a r_1 a r_2... a r_k \mod m$$
Where $k=\phi(m)$. I don't understand this step, please explain.
 A: The function multiply-by-$a$-mod $m$ is a bijection, ie. a permutation on the set $\{r_1,r_2,\ldots, r_k\}$. SO the product on either sides in your question are equal mod m.  To check it is a bijection suffices to check it is one-one.
Suppose $ar_i\equiv ar_j \pmod m$. This shows $m$ divides $a(r_i-r_j)$. But as $\gcd(a,m)=1$ it follows that $r_i-r_j$ must be a multiple of $m$. As all the $r_i$'s are less than $m$ this is possible only when the difference is zero, hence $r_i=r_j$. QED
A: A reduced residue system mod $m$ consists of those congruence classes that are relatively prime to $m$. Stated differently, we say that $r_1, \ldots, r_k$ is a reduced residue system mod $m$ if it is a reordering of every number (mod $m$) from $1$ to $m-1$ that is relatively prime to $m$.
So the two lists of numbers $r_1, r_2, \ldots, r_k$ and $ar_1, ar_2, \ldots, ar_k$ are the same lists of numbers (mod $m$, that is). Therefore multiplying them together mod $m$ must give the same product, which is the step you are asking about.
Let's do an example. Suppose we're interested in showing $2^6 \equiv 1 \pmod 9$. Then a reduced residue system mod $9$ is $$1,2,4,5,7,8.$$
What happens when we multiply each of these by $2$? At first, we get
$$ 2,4,8,10,14,16,$$
and after reducing mod $9$ we get
$$ 2,4,8,1,5,7.$$
Notice that this is exactly the same as the first list, but in a different order.
Since these are the same lists, their products must be equal. Here, that means that
$$ 1\cdot 2 \cdot 4 \cdot 5 \cdot 7 \cdot 8 \equiv 2 \cdot 4 \cdot 8 \cdot 10 \cdot 14 \cdot 16 \equiv 2^6 \cdot 1 \cdot 2 \cdot 4 \cdot 5 \cdot 7 \cdot 8 \pmod 9.$$
After multiplying by the modular inverse of $1 \cdot 2 \cdot 4 \cdot 5 \cdot 7 \cdot 8$, we see that
$$ 2^6 \equiv 1 \pmod 9.$$
Do you see how it works now?
