In characteristic $p$, every irreducible polynomial is uniquely $g(x^{p^e})$ with $g$ irreducible and separable

The derivative $f^{\mathrm{\prime }}$ is zero exactly when every exponent occurring in $f$ is divisible by $p$; in that case there is a unique $h\in F[x]$ with $f(x)=h(x^p)$. Repeating this finite descent in degree gives a unique maximal $e$ and a polynomial $g$ with $f(x)=g(x^{p^e})$ and $g^{\mathrm{\prime }}\mathrm{\ne }0$.

If $g=uv$ with both factors nonconstant, then $f=u(x^{p^e})v(x^{p^e})$, contradicting irreducibility of $f$; hence $g$ is irreducible.

Since $g^{\mathrm{\prime }}\mathrm{\ne }0$, any nonunit common divisor of $g$ and $g^{\mathrm{\prime }}$ has an irreducible factor by [L3], which would divide the irreducible $g$ and hence force $g\mid g^{\mathrm{\prime }}$, impossible by degree; thus $\gcd (g,g^{\mathrm{\prime }})=1$ and [L1] makes $g$ separable.

The exponents occurring in $f$ determine their largest common power $p^e$, so $e$ and then the coefficient-preserving core $g$ are unique. Moreover $e=0$ exactly when $f^{\mathrm{\prime }}\mathrm{\ne }0$, which for irreducible $f$ is equivalent to separability by [L1].