Multiplicity of a root of a polynomial :)

It's true that, if a polynomial has a root (let's say,  k, for example) with multiplicity n (n>1, for n integer), then it's true that the derivate polynomial have k as a root with multiplicity n-1.

The reciprocal is always true? I.e., it's true that if a derivative polynomial have k as a root with multiplicity n-1, then the original polynomial have k as a root with multiplicity n? If this is not always true, in which conditions this is true?

Thanks every one! :)
The complete story is this:

The following conditions are equivalent for a polynomial $f(x)$:

(i) $\;f(x)$ is divisible by $(x-a)^n$;

(ii) $\;f(a)=f'(a)=\dots=f^{(n-1)}(a)=0$.

It is a generalisation of the well-known result:

$f(x)$ has $a$ as a root  if and only if $f(x)$ is divisible by $x-a$.
suppose $f$ is a polynomial of degree $m$ and $f'$ has a root $k$ of multiplicity $k.$ then we can write $$f'(t) = (t-a)^{n-1}q(t), q(a) \neq 0.  \tag 1  $$ 
integrating $(1)$ from $a$ to $x,$ we get 
$$\begin{align}f(x)-f(a) &amp;= \int_a^x(t-a)^{n-1}q(t)\, dt \\
&amp;=\int_0^{x-a}t^{n-1}q(t+a)\, dt\\
&amp;=\int_0^{x-a} t^{n-1}\left(q(a)+a_1t+a_2t^2 +\cdots+a_{m-n+1}t^{m-n+1}\right)\, dt\\&amp;=(x-a)^n\left(\frac{q(a)}n+\frac{a_1}{n+1}(x-a)+\cdots+a_{m-n+1}(x-a)^m\right)\end{align}$$ therefore if $f(a) = 0,$  then the zero $x=a$ of $f$ has multiplicity $n.$