Theorem 1
If p is prime, then p divides 
 n(p-1)      n(p-1)
a       -   b       for integers (a,b)=1,n p where p does not divide ab. 
No matter what the value of n, there exist a1,b1 such that a1 = a^n and b1=b^n.
Thus, we can write theorem I as

If p is prime and (a1,b1) = 1, then a1^(p-1) - b1^(p-1) is divisible by p.

Some examples in origional form.
a = 3 b = 2
 n(p-1)   n(p-1)
a      - b
n p
--- ---- --------------------
1 7 5*7*19
2 7 5*7*13*19*61
1 107 5*13*107*...
7 107 5*29*71*107*...

Theorem 1a (Euler) (a,m) = 1 => a^Phi(m) - 1 is div by m.
Phi(m) = number of non divisors <= m of m. If m is prime,
Phi(m) = p-1.

Lemma 1
Let b = 1. Prove if p is prime and (a1,p) = 1, a1^(p-1) - 1 is div by p.
We know from Theorem 1a, (a,m) = 1 => a^Phi(m) - 1 is div by m.
Then a1^(p-1)-1 = a^Phi(p)-1 since p is prime. So p divides a^(p-1) - 1.

Let a = k^n for some n
By rules of exponents, k^n(p-1) - 1 = (k^n)^(p-1) - 1 = k^n(p-1). Thus, 
letting a = k^n we get the desired result.

We can now prove Theorem 1.
by Lemma 1 we can assume a and b are powers of n and
a^(p-1)-1 is div by p and b^(p-1)-1 is div by p then
a^(p-1) - 1 - b^(p-1) - 1 = a^(p-1) - b^(p-1) is div by p as desired.