Here are several different ways of stating Euler's theorem. The first is how I remember it.
summary: modular exponentiation works, except that
version 1: In other words, if [a]∈ℤm is a unit and [b]∈ℤϕ(m) then [a][b] = [ab] is well defined.
version 2: if [a]∈(ℤm)* and b ≡ b′ (mod ϕ(m)) then [a]b = [a]b′.
version 3: if [a] is a unit of ℤm, then [a]b + kϕ(m) = [a]b (mod m)
version 4: if [a] is a unit of ℤm, then [a]ϕ(m) = [1] (mod m)
It is a good exercise to check that these are all saying the same thing. Version 1 is closest to the summary, and is the easiest to use in practice. Version 4 is easiest to prove. Versions 2 and 3 help to show how version 1 and 4 are equivalent.
Summary: We wish to show that [a]ϕ(m) = [1]. We will draw a picture of what happens when you multiply all the units by [a]. We'll find that the picture always forms k loops of the same size (n), so that there are kn elements (i.e. ϕ(m)=kn). To find [a]x, you start at [1] and multiply by [a] x times, which traverses x arcs. If x = ϕ(m) then this goes around a loop exactly k times, ending on [1].
Let's consider m = 9. Then (using the fact that [a] is a unit if a and m are coprime):
(Zm)* = {[1],[2],[4],[5],[7],[8]}, ϕ(9)=6.
We draw an arc from x to y if [a]x = y. For example, if a = 2, then we have
[1] --> [2] --> [4] --> [8] --> [7] --> [5]
| | multiplication by [2]
| | ------------------------>
\-------------------<<<-----------------/(note that [8][2]=[16]=[7] and [7][2]=[14]=[5] mod 9). We get a nice loops that goes through all of the units. To compute [2]6, we start at [1] and multiply by [2] 6 times, i.e. we start at [1] and follow 6 edges. This brings us back to [1], so [2]6 = [1].
We get a different picture if a = 4:
[1] --> [4] --> [7] [2] --> [8] --> [5]
| | | | multiplication by [4]
\-----<<<-------/ \------<<<------/ ------------------------>Here we have two loops, both of size 3. To compute [4]6, we start at [1] and follow 6 edges. This goes around the first loop twice, ending up at [1]. Thus [4]6 = [1].
We get a third picture if a = 1:
[1] [2] [4] [5] [7] [8]
/ \ / \ / \ / \ / \ / \ multiplication by [1]
\-<-/ \-<-/ \-<-/ \-<-/ \-<-/ \-<-/ ------------------------>Here we have 6 loops, each of size 1. We compute [1]6 by starting at [1] and following 6 edges. This loops 6 times around the loop of size 1, ending at [1], so [1]6 = [1].
In general, the picture will always break up into k loops, all of the same length. There are several things you may need to check to convince yourself of this. They all have very simple (one or two line) proofs, which I encourage you to check. I have done the first for you:
Proof: There can't be two (different) edges out of x, say to y and z, because otherwise we have [y]=[a]x = [z].
There also can't be two edges (say from y and z) into x, because otherwise we have [a]y = x = [a]z. Multiplying both sides by [a]−1 (which exists because [a] is a unit) gives y = [a]−1[a]y = [a]−1[a]z = z
The arcs stay within (Zm)*. In other words, the product of units is a unit.
If n is the length of the loop containing [1] (and [a]), then for any [b], traversing n edges from [b] leads back to [b].
Similarly, traversing fewer than n edges does not lead back to [b].
Thus all of the loops are the same size (call it n), and every element is in exactly one of the loops. So the loops split up (Zm)* into k groups of n elements each, so (Zm)* has kn elements. In other words, ϕ(m)=kn.
Now, to compute [a]ϕ(m), we start at [1] and multiply by [a] ϕ(m) times. In other words, we traverse kn arcs. This means we go around the loop containing [1] exactly k times, so we end up back at [1]. Thus [a]ϕ(m) = [1].