Want to show a finite field has order p^n, where p is prime

ATTEMPT:

Let F be a field, then char(F) = p, a prime (Using some theorems in my algebra book). So p*1 = 1+1+...+1=0 so |1|=p. Then for any r in F p*r = r+r+...+r=r*(1+1+...+1)=0 implies |r| = p. so p divides |F|

Now assume that q is a prime, q not equal to p and q divides |F|. By Cauchy's theorem for finite abelian groups, F then has an element of order q, which is a contradiction that all elements have order p.

I do not know where I can go from there...Can I just say that |F| = p^n? I do not see that. THANKS!!!