Please I would like you to tell me if my proof is correct
Let $f(x)\in K[x]$ ($K$ field). Prove that if $(f(x),f´(x))=1$ (greates common divisor is 1) then $f(x)$ does not have multiple roots in $K$
my attempt: Suppose that $f(x)$ has multiple roots in $K$. Let $a$ be one of those roots with multiplicity $m$ then $f(x)=(x-a)^{m}g(x)$ for some $g(x)\in K[x]$.
If $a$ is a root of $f(x)$ with multiplicity $m$ then $a$ is a root of $f´(x)$ with multiplicity $m-1$ hence $f´(x)=(x-a)^{m-1}h(x)$ for some $h(x)\in K[x]$
$$f(x)=(x-a)^{m}g(x)=(x-a)^{m-1}(x-a)g(x)=(x-a)^{m-1}q(x)$$ ($q(x)=(x-a)g(x)$) therefore
$(x-a)^{m-1}|f(x)$ ($(x-a)^{m-1}$ divides $f(x)$) and
$(x-a)^{m-1}|f´(x)$ ($(x-a)^{m-1}$ divides $f´(x)$)
but this is a contradiction because $(f(x),f´(x))=1$
therefore $f(x)$ does not have multiple roots in $K$
I really would appreciate your help :)