Let $p$ be a prime number and $n\in\mathbb N$. Consider the determinant
$$M_n = \begin{vmatrix}\frac1{x^{p^{n+1}}-x}&\frac1{x^{p^{n+1}}-x^p}\\ \frac1{x^{p^{n+2}}-x}&\frac1{x^{p^{n+2}}-x^p}\end{vmatrix} \in \mathbb F_p(x)$$
Numerical computations suggest that
$$\deg(M_n)=p-(p+2)p^{n+1}$$
Is it true? Is yes, does anyone have an idea to prove it?
Assuming that by degree you mean degree of the numerator minus the degree of the denominator, this is a trivial calculation. The determinant is $$\frac{1}{(x^{p^{n+1}}-x)(x^{p^{n+2}}-x^p)}-\frac{1}{(x^{p^{n+2}}-x)(x^{p^{n+1}}-x^p)}.$$ Multiplying the denominators of these fractions to combine them gives a denominator of degree $2(p^{n+1}+p^{n+2})$. In the combined numerator, the $x^{p^{n+1}+p^{n+2}}$ terms will cancel, leaving the highest degree term in the numerator as $x^{p^{n+2}+p}$. So the degree of the numerator minus the degree of the denominator is $$(p^{n+2}+p)-2(p^{n+1}+p^{n+2})=p-(p+2)p^{n+1}.$$