Is there a known proof without differentiating that proves that all irreducible polynomials over $\mathbb{Q}$ are separable? (Or even better, for all fields of characteristic $0$.)
EDIT: As people seem to question this thread; I do know a proof with derivatives - my motivation for one without is simply curiosity. Multiple approaches are always nice.
Let $L$ be the splitting field of $f \in F[x]$. By elementary properties of field extensions, the automorphism group of $L/F$ acts transitively on the roots, so they have equal multiplicity. Suppose wlog that $f$ is monic. Then $f$ is some $n$th power of a $g \in L[x]$. By inspection of coefficients, $g^n \in F[x]$ implies $g \in F[x]$. Contradiction.
In the last step, it is used that $n \neq 0$ in F.
More details for that step: Let $d$ be the degree of $g$. We show by induction that the coefficient $a_{d-i}$ of $x^{d-i}$ in $g$ lies in $F$. For $i = 0$ this is clear. Suppose true for $0, \ldots, i-1$, and look at the coefficient of $x^{nd-i}$ in $g(x)^n$. It equals $n a_{d-i}$ $+$ a polynomial expression involving the $a_{d-j}$ for $j < i$. Because $n \neq 0$ and those $a_{d-j} \in F$ by assumption, we have $a_{d-i} \in F$.