Proof. sketch [1702A]

WLOG, assume $f(x)$ is monic in $F[x]$.

Let $\alpha, \beta \in \overline{F}$ be roots of $f(x)$ with multiplicities $a, b$ respectively. Suppose $a < b$. Since $f(x)$ is monic and irreducible in $F[x]$, $m_{\alpha, F}(x) = m_{\beta, F}(x) = f(x)$. Therefore, by Theorem 1408, there exists a field isomorphism $$ \varphi : F(\alpha) \to F(\beta) $$ such that $\varphi(\alpha) = \beta$ and $\varphi(r) = r$ for all $r \in F$.

Over $F(\alpha)$, $f(x) = m_{\alpha, F}(x) = (x - \alpha)^{a}g(x)$ for some $g(x) \in F(\alpha)[x]$ such that $(x - \alpha) \nmid g(x)$. Over $F(\beta)$, $f(x) = m_{\beta, F}(x) = (x - \beta)^{b}h(x)$ for some $h(x) \in F(\beta)[x]$ such that $(x - \beta) \nmid h(x)$. Then $$ \begin{align*} \varphi \left( f(x) \right) &= (x - \varphi(\alpha))^{a} \varphi \left( g(x) \right) \\ &= (x- \beta)^{a} \varphi \left( g(x) \right) \\ &= (x - \beta)^{b}h(x). \end{align*} $$ It follows that $\varphi \left( g(x) \right) = (x - \beta)^{b-a}h(x)$. Since $b > a$, $(x - \beta) \mid \varphi \left( g(x) \right)$. Apply $\varphi ^{-1}$ on $\varphi \left( g(x) \right)$, it turns out that $$ g(x) = \varphi ^{-1} \varphi \left( g(x) \right) = (x - \alpha)^{b-a} \varphi ^{-1} \left( h(x) \right). $$ Hence $(x - \alpha) \mid g(x)$ in $F(\alpha)$, contradicting the fact that $(x - \alpha) \nmid g(x)$. So we must have $a = b$, i.e. the multiplicities of $\alpha$ and $\beta$ are equal. Because $\alpha, \beta$ are arbitrary, all roots of $f(x)$ in $\overline{F}$ have the same multiplicity.