Theorem 1.
Let N = (p+1)^p - p^p   f(i) be the ith divisor of N. Then if p is prime
f(i) == 1 mod p for all i.

Corollary 1.
If p is not prime then the prime divisors of p divide f(i) -1 for one or
more i.

From algebra we have
(1)  a^n - b^n = (a-b)(a^(n-1) + a^(n-2)b + . . . + b^(n-1). 

Certainly if p is prime and N is prime the theorem is true since
from (1) a = p+1 and b = p so that
N    =  ((p+1)^(p-1) + (p+1)^p-2p + . . . + p^p-1)  and expanding,
N-1 =  pA for some integer A.