Let $f(x)\in F[x]$, degree$(f(x))=n>1$, char$(F)=p$, prime and $f'(x)=0$. Prove, $p\mid n$ and $f(x)$ has at most $n/p$ distinct roots.

I have just proved the first part $$p\mid n$$ in the following manner.
Let $$f(x)=a_0+a_1 x+\cdots+a_n x^n$$ where $$a_n\ne0$$
Now, $$f'(x)=a_1+2a_2 x+\cdots+na_n x^{n-1}=0\implies ia_i=0\ \forall i=1, 2,\ldots,n$$
$$\implies$$either $$p\mid i$$ or $$a_i=0\ \forall i=1, 2,\ldots,n$$
In particular since $$a_n\ne 0$$, $$p\mid n$$(Done)
But I can't prove the second part of this problem that is $$f(x)$$ has at most $$n/p$$ distinct roots.
My thought is- if I can prove any root of $$f(x)$$ has multiplicity at least $$p$$, then we are done. But I can't prove it. I even don't know whether my intuition is correct or not.
Can anybody solve the second part? Thanks for assistance in advance.

Under these conditions, your $$f(x)$$ has the form $$f(x)=a_0+a_px^p+a_{2p}x^{2p}+\cdots+a_n x^n.$$ Let $$b_j$$ be a $$p$$-th root of $$a_{pj}$$ in some extension field of $$F$$. Then $$f(x)=g(x)^p$$ where $$g(x)=b_0+b_1x+b_2x^2+\cdots+b_{n/p}x^{n/p}.$$ This is now defined over an extension field of $$F$$, not necessarily $$F$$ itself, but no matter, each zero of $$f$$ is a zero of $$g$$, and $$g$$ has $$\le n/p$$ zeroes.
• Ok, but at the last line I think Every Zero of $g$ is a zero of $f$, since if $g(\alpha)=0\implies g(\alpha)^p=0\implies f(\alpha)=0$ – Biswarup Saha Oct 26 '18 at 5:38
• Both $\implies$s are $\iff$s. @BiswarupSaha – Lord Shark the Unknown Oct 26 '18 at 6:27
• Lord Shark the Unknown, one moment why such $b_j$ will always exist. If $F$ is finite or perfect then okay, but else? – Biswarup Saha Dec 5 '18 at 14:38
• @BiswarupSaha "...in some extension field of $F$." – Lord Shark the Unknown Dec 5 '18 at 15:04