 Solution

For nonconstant irreducible $f$, its greatest common divisor with $f^{\prime }$ is either $1$ or $f$. Since $\mathrm{deg}{f}^{\prime }<\mathrm{deg}f$ when $f^{\prime }\ne 0$, inseparability is equivalent to $f^{\prime }=0$. In characteristic $p$,exactly when $a_j=0$ for every exponent $j$ not divisible by $p$. ThusNo assumption that the coefficient field is perfect is required: vanishing derivative concerns the exponents, not whether the coefficients have $p$th roots in $F$.