Let $p$ be a prime number. Consider for all $k,r\in\mathbb{N}_0$ with $k|r$ the field $\mathbb{F}_{p^k}$ as a subfield of $\mathbb{F}_{p^r}$. Define
$$ \overline{\mathbb{F}_p}:=\bigcup_{i\in\mathbb{N}_0}\mathbb{F}_{p^i}.$$
I want to show that $\overline{\mathbb{F}_p}$ is the algebraic closure of $\mathbb{F}_p$. I already managed to show that $\overline{\mathbb{F}_p}$ is a field and that it is algebraic over $\mathbb{F}_p$, but I can't seem to show that it is also algebraically closed.
I know that for every $i\in\mathbb{N}_0$, $\mathbb{F}_{p^i}$ is a decomposition field of the polynomial $X^{p^i}-X$. Which means this polynomial can be expressed as $$X^{p^i}-X=\prod_{a\in\mathbb{F}_{p^i}}(X-a),$$
and I thought that this could help me prove that every irreducibel polynomial in $\overline{\mathbb{F}_p}[X]$ has degree $1$ but so far it hasn't.
Any help would be appreciated.
If $q(x) \in \mathbb{F}_{p^k}[x]$ is of degree $n$ and irreducible, $q$ has a non zero root in the rupture field $\mathbb{F}_{p^k}[x]/(q)$ which is isomorphic to $\mathbb{F}_{(p^k)^n}= \mathbb{F}_{p^{kn}}$.
By induction, you can prove that $q$ splits in one of the $\mathbb{F}_{p^m}$. Which implies the desired result.