Let $E=F[\alpha]$ be a proper extension of fields and $\alpha^{p^{n}}\in F$ for $p= \text{char}(F)$ and some integer $n > 0$. Show that there exists a $F$-derivation $\delta$ on $E$ for which $\delta(\alpha) = 1$.
Where an $F$-derivation is a linear (with respect to scalar multiplication from $F$) map $\delta:E\to E$ which obeys the product rule in $E$: $$\delta(xy) = x\delta(y) + y \delta(x)$$.
The textbook gives a hint to define $\delta$ on the set $1,\alpha,\alpha^{2},\ldots,\alpha^{m-1}$ where $m$ is the degree of the extension. I see what to do for that, just the usual calculus derivative. This set forms a basis for $E$, so then extend by linearity to all of $E$. We then just have to check the product rule. I believe it just suffices to check that $\alpha^{k}=k\alpha^{k-1}$ for all $k$ (not just $k < m$). It seems like it should just be calculation, but I am struggling this way and I don't see where the assumption $\alpha^{p^{n}}\in F$ will be used.
This is from Martin Isaacs "Algebra: A Graduate Course".
I think I figured it out.
Notice that $m=p^{n}$, the extension $E$ is purely inseparable and the minimal polynomial of $\alpha$ is $x^{p^{n}}-\alpha^{p^{n}}$. Because $\delta$ is $0$ on $F$ we have $$ \delta(\alpha^{p^{n}}) = 0 = p^{n}\alpha^{p^{n}-1} $$ as needed. For any $k$ we can write $k=qp^{n}+r$ with $0\leq r < p^{n}$ so that $$\delta(\alpha^{k}) =\alpha^{qp^{n}}\delta(\alpha^{r}) = r\alpha^{qp^{n}+r-1} = k\alpha^{k-1} $$ because $r\cong k \mod p$