Before anyone closes this question thinking that SE already has an answer to this question, please read my post below(if it still answers it somewhere, let me know and I will remove it).
My proof: The standard proof constructs a new polynomial like
$\prod_{\alpha \in F} (x-\alpha) + 1$ and arrive at a contradiction. However, since this didn’t strike me at first so the following is what I tried.
Let $F$ be an algebraically closed field and if possible, suppose it is finite. Then char$(F)=p$, a prime and in fact, $|F| = p^n$ for some positive integer $n$. Consider now the polynomial $p(x)= (x^{p^n})^m-x = x^{p^n}\dots x^{p^n} -x = x^{m. p^n}-x \in F[x]$ for some integer $ m >1$. Let $K$ be the splitting field of $p(x)$.
Now note that since $p’(x) = -1$, so $p(x)$ is separable over $F$ and thus it has $mp^n$ distinct roots in some splitting field(say, $K$ with $K \supset F)$ of $p(x)$. This implies that $|K| \geq mp^n > p^n =|F|$. So $ K \neq F$ and so we have found a polynomial $p(x) \in F[x]$ which does not have all its roots lying in F, an algebraically closed field. This is a contradiction.
Can someone check if the above argument is true? I would appreciate any output any of you have on this proof.
Thank you very much.