Finite fields are isomorphic

This is from A Course in Arithmetic by JP Serre

Theorem 1

ii) Let $p$ be a prime number and let $q = p^f(f \geq 1)$ be a power of $p$. Let be an algebraically closed field $\Omega$ of characteristic $p$. There exists a unique subfield $F_q$ of $\Omega$ which has $q$ elements. It is the set of roots of the polynomial$X^q-X$.

iii) All finite fields with $p^f$ elements are isomorphic to $F_q$.

The proof of the last part says

Assertion iii) follows from ii) and from the fact that all fields with $p^f$ elements can be embedded in $\Omega$, since $\Omega$ is algebraically closed.

Can someone explain me how iii) follows fom ii)?

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If we had two fields $L,K$ which have cardinality $p^f$, then both are algebraic over $\mathbb{F}_p$, and so they both embed into $\overline{\mathbb{F}_p}$, but by (ii) they have the same image, and in particular are both isomorphic to that image and thus to each other. – John Stalfos Jan 30 '13 at 14:12

Let $E$ be a field with $q$ elements. By (iii), it is isomorphic to a subfield of $\Omega$. By (ii), there's only one such subfield, namely $F_q$.
You can also use a little Galois Theory: by (ii) any field $\,\Bbb F_q\,$ is the splitting field of $\,x^q-x\,$ over $\,\Bbb F_p\,$ , and "the" comes from the uniqueness, up to isomorphism, of the splitting field...