Finite fields, existence of field of order $p^n$,proof help I have to prove the following theorem :

Let $p$ be a prime number and let $n \ge 1$,be any integer, then there exists a field of order $p^n$.

My attempt
I started off by considering the polynomial $f(x)$=$x^{p^n}-x \in \Bbb Z_p[x]$.
I took $F$ to be the splitting field of $f(x)$ over $\Bbb Z_p$.
Since $F$ is the splitting field of $f(x)$ over $\Bbb Z_p$ therefore $f(x)$ has exactly $p^n$ zeros in $F$ counting multiplicity.
Also since $f'(x)$ = $p^nx^{p^n-1}-1$ , therefore $f(x)$ will not have any multiple zeros and hence it will have $p^n$ distinct zeros in $F$.
Then I took $K =\{k \in F | k^{p^n} = k\}$.
I was trying to prove that $K$ is a sub-field of $F$.
$1 \in K$ thus $K$ is non-empty.
For $a,b \in K$ $a^{p^n}=a$ and  $b^{p^n}=b$
So $(a+b)^{p^n}$ (should be) = $a^{p^n}$+$b^{p^n}$, this is where I am currently stuck!
I know that this is true when  $p$ is the characteristic of a field, but why would this expansion be true here? 
 A: In $\mathbb Z_p$ and hence its extension $F$ we have
$$
p a = \underbrace{a + \cdots + a}_{p \text{ times}} = (\underbrace{1 + \cdots + 1}_{p \text{ times}})a = 0 \cdot a = 0.
$$
By the binomial theorem
$$
(a + b)^{p^n} = \sum_{k=0}^{p^n}\binom{p^n}{k} a^k b^{p^n - k}.
$$
All terms in the latter expansion are multiples of $p$ except for $a^{p^n}$ and $b^{p^n}$. It follows that in $\mathbb Z_p$ and $F$ we have
$$(a + b)^{p^n} = a^{p^n} + b^{p^n}.$$
Now show that the roots of $f(x)$ are closed under addition, multiplication and taking additive and multiplicative inverses. This shows that $F$ consists entirely of the roots of $f(x)$. We conclude that the order of $F$ is $p^n$ as desired.
A: We note that it is sufficient to show there exists an irreducible polynomial, $Q(x)$ of degree $n$ over $\Bbb F_p$, the field with $p$ elements (which you denote by $\Bbb Z_p$). This is so because in this case you have that $\Bbb F_p[x]/(Q(x))$ is a field of degree $n$ over $\Bbb F_p$, hence has $p^n$ elements.
Consider the polynomial $x^{p^n}-x$, which has no repeated roots, since its derivative is $-1$. Then I claim the splitting field of this polynomial exists and has degree $n$ over $\Bbb F_p$. For each divisor $d | n$ with $d < n$, consider the polynomial $\Phi_d(x)$ which we define to be the product of all irreducible degree-$d$ polynomials over $\Bbb F_p$. Then define
$$P(x) = \left(x^{p^n-1}-1\right) \bigg/ \prod_{d|n,\; d<n} \Phi_d(x)$$
Since all the roots of $x^{p^d}-x$ are in one of the factors we've factored out, the representation of $P(x)$ as a product of irreducibles satisfies the fact that every root of every irreducible factor has multiplicative order exactly $p^n$. Choose one such root, $\omega$. Then we have that $\deg \omega\ge n$ because $x^{p^d}\ne 1$ for $d<n$. However we also see that the $\deg \omega \le n$ because if it generated an extension of degree $k<n$, we can use the fact that the multiplicative group of a field is a group, and we'd have $x^{p^k}=1$, a contradiction. So $\deg\omega =n$, and if $m_\omega(x)$ is it's minimal polynomial we have that $m_\omega(x)=Q(x)$ is the desired polynomial.
