Of course r need not be irreducible, but u is a root of some irreducible polynomial q dividing r, and Zp(u) includes all the roots of q, giving a normal, simple extension. Although q is not known, its degree is n, since this is an n dimensional extension.
Suppose there are two such extensions. They both split r(x), and they both split q(x), the irreducible factor described above. Select u in the first extension and v in the second, both roots of q(x). Adjoining u or v creates the entire extension, of dimension n, since q has degree n. Let c() fix Zp, and carry u onto v. This builds an isomorphism between the two extensions. Thus the finite field of order pn is unique.
We know what the field looks like if it exists; let's prove it exists. Given p and n, build r(x) as above, and let F be the splitting field for r(x) over Zp. In other words, adjoin roots of r(x) until F splits r(x). Let's show that all these roots are distinct.
If r(x) has any repeated roots, they will show up in the formal derivative. Yet the derivative is a power of x, with roots of 0, hence r(x) has distinct roots. Our splitting field F has at least pn elements.
Let c be the frobenius homomorphism c(x) = xp. A ring homomorphism on a field is injective, and since both sets are finite, it is also surjective, hence a field automorphism.
Let d = cn. Thus d is an automorphism, the result of applying c n times. Let L be the subfield of F that is fixed by d. Every x in L satisfies xpn = x. This is precisely 0 and the roots of r(x). Therefore L is a finite field of order pn.