Take the following proof that every algebraically closed field is infinite.
Suppose we have a finite, algebraically closed field $F$. The polynomial $$P\left(X\right)=\prod _{f\in F}\left(X-f\right)+1$$ has no roots. It is clear that $P\in F\left[X\right]$, and that $P$ has no roots. However, to show that $F$ is not algebraically closed, we need a polynomial with degree at least $1$. Isn't this polynomial degree $0$? It is equal to $1$ identically?
Alternatively, if a field $F$ is finite and has $n$ elements, then all elements of $F$ are roots of $x^n-x$ and so the polynomial $x^n-x+1$ has no roots in $F$. Thus, $F$ is not algebraically closed.
This is essentially the same proof, since $x^n-x=\prod _{f\in F} (x-f)$, but perhaps it's psychologically clearer.