## If a is algebraic over F of degree at least 2, then no power of x-a is in F[x]

Let $F$ be a field and let $\alpha$ be algebraic over $F$ of degree at least 2. Prove that $(x-\alpha)^n$ is not in $F[x]$ for any $n$.

Suppose to the contrary that $(x-\alpha)^n \in F[x]$. Then the minimal polynomial $p(x)$ of $\alpha$ over $F$ divides $(x-\alpha)^n$. Since the roots of $(x-\alpha)^n$ are all the same and those of $p(x)$ are all distinct, in fact $p(x) = x - \alpha$; but then $\alpha$ has degree 1 over $F$, a contradiction.

So no power of $x-\alpha$ can be in $F[x]$.