# Prove that no finite field is algebraically closed (using a particular method)

I'll preface this question by saying this question is not a duplicate as I am looking for insight into proving this a particular way. I have found a proof of the result both here on stackexchange and elsewhere.

This question, in particular, is from $\S6.4$ of W. Keith Nicholson's "Introduction to Abstract Algebra". It says:

Show that no finite field F is algebraically closed. [Hint: Apply Exercise 17 to $f = x^{q+1}+1$ where $q = |F|$.

Here, Exercise 17 says:

Let $f$ be a nonconstant polynomial in $F[x]$. Show that $f$ has no repeated root in any splitting field of $F$ if and only if $f$ and $f'$ are relatively prime in $F[x]$.

So, in this case $f = x^{q+1} + 1$ gives us that $f' = (q+1)x^q$. But I'm not sure what the relationship between any of these things is.

Any guidance would be appreciated.

If conversely your finite field $F$ is algebraically closed, then $f$ splits into linear factors. As a polynomial of order $q+1$, $f$ has at least one repeated root. By your ex.17, $f$ and $f'$ are not relatively prime, which is a contradiction because they are:
$f - \frac{xf'}{q+1} = 1$, so any common divisor is a divisor of $1$.