Order of finite fields is $p^n$

Let $F$ be a finite field. .How do I prove that the order of $F$ is always of order $p^n$ where $p$ is prime?

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What is the cardinality of any finite-dimensional vector space over the field with $p$ elements? –  NKS Oct 15 '11 at 17:50
What exactly is the scalar field and the multiplication law? –  Mohan Oct 15 '11 at 18:14
The scalar field is the subfield consisting of the elements $0$, $1$, $1+1$, $1+1+1$, $1+1+1+1$ and so forth. (You first have to prove this is in fact a subfield, of course). Multiplication is whatever passes for multiplication in the finite field. –  Henning Makholm Oct 15 '11 at 18:37
First prove that if $charF=p$ then $F_p$ is a subfield of $F$. –  Frank Murphy Oct 15 '11 at 19:27
The answers to this question contain all the information that you need. IOW this is almost an exact duplicate. –  Jyrki Lahtonen Oct 16 '11 at 6:17
1. Prove that the smallest multiple $m$ of 1 that gives zero has to be a prime. (Otherwise there are divisors of $m$ which are then divisors of zero.)
3. Count the elements of the field if the dimension of this vector space is $n$.