I have to show that all field with $p^2$ ($p$ prime) elements are isomorphic to each other. Let $F$ be any such field. Since $F$ has characteristics $p$, it contains a prime field $\Bbb F_p$, so it can be said that $F$ is quadratic extension of $\Bbb F_p$. Let $x =a + b \alpha$ where $a, b \in \Bbb F_p$ and $\alpha \in F$. Then, by Frobenius endomorphism $$x^{p^2} - x = 0$$ so elements of $F$ are elements of $\Bbb F_{p^2}$ as well. Every field with $p^2$ elements is isomorphic to $\Bbb F_{p^2}$. Thus there exists one up to isomorphism.
Is my reasoning correct? If not, I would be grateful some hints. There is another method in this book (page 210) and problem is from page no 57. I just don't understand it and I don't see why my idea is wrong or right.
Suppose that $\mathbb{F}$ is a finite field such that $|\mathbb{F}|=p^2$.
It is not difficult to show that $a^{\,p^2}=a \,\,\,\forall a\in \mathbb{F}$.
Let $\mathbb{F}_p\cong \mathbb{Z}_p$ the prime field of $\mathbb{F}$, and $f(x)=x^{p^2}-x \in \mathbb{F}_p[x]$. Since $f(x)$ has at most $p^2$ roots and $f(a)=0\,\, \forall a \in \mathbb{F}$, then $\mathbb{F}$ is a splitting field of $f(x)$ over $\mathbb{F}_p$.
Now you can use the fact that any two splitting fields of a polynomial $f(x)$ over the same field are isomorphic.