Let the set S contain p-tuples of elements of G, such that the product of the p entries is always 1. Thus |S| = |G|p-1.
The group Zp acts on S by performing right circular shifts. Remember that orbit length times stabilizer size is p, and since p is prime, the only choices are 1 and p. If every orbit other than 1.1.1… has size p, |S| is congruent to 1 mod p, which is impossible. Therefore some nontrivial tuple is its own orbit, and is fixed by all circular shifts. All elements in this tuple are equal, hence xp = 1. In other words, x generates a cyclic subgroup of order p inside G.
In fact, the number of elements of order p is congruent to -1 mod p. Combine this with 1.1.1…, and the orbits of size p, and build all the tuples in S, with |S| divisible by p. This can be satisfied by precisely one p subgroup, having p-1 elements of order p. Or you might have lots of elements of order p.