Proving that, if a polynomial with real coefficients has all real roots, then the roots of its derivative are real, too. I need to prove that

The roots of the derivative of a polynomial $f$ of degree $n$ (with real coefficients) whose roots are all real are real, too.

The hypothesis of the roots of $f$ being all real implies that we can express $f$ as a product
$$a(x-x_1)\cdots (x-x_n).$$
Let's define $f_i(x):= x-x_i$. My idea was to calculate the derivative of $f$ applying the general Leibniz rule to it's decomposition above, and to somehow obtain from it an expression of $f'$ as a product of linear polynomials (concluding
this way our proof).
If I'm not wrong, since all the multinomial coefficients equal 1 and the derivative of each function $f_i$ is one, we obtain that
$f'=a(f_1\cdots f_n)'=a[ f_2 \cdots f_n + f_1 f_3 \cdots f_n+ \cdots + f_1f_2 \cdots f_{n-1}]$.
Now I have no ideas of how to factorize it as a product of the $f_i's$ (or in general as a product of linear factors). Any suggestion is very appreciated.
 A: With the useful point of view of @dan_fulea in the comments I was able to prove the statement, I'm sharing my development.
Let $f$ be a polynomial with real coefficients whose roots $x_1, \ldots , x_m$ are all real. If we denote the multiplicity of the root $x_i$ by $r_i$, then
$$f(x)=a(x-x_1)^{r_1}\cdots (x-x_m)^{r_m}.$$
Let's suppose further that just the first $l$ roots of $f$ have multiplicity greater than 1. Then, $$\delta(f)=r_1+ \cdots +r_l +(m-l).$$
Note first that, by Role's theorem, we know about the existence of $m-1$ roots $\xi_i$ of the polynomial $f'(x)$, with $\xi_i \in ]x_i, x_i+1[$ for all $i$. Now, we assert that for each $1 \leq i \leq l$, the number $x_i$ is a root of $f'$.
Indeed, expressing $f$ as the product $f(x)=(x-x_i)^{r_i}g(x)$, by the rule of the derivative of a product we get that
$$f'(x)=r_i(x-x_i)^{r_i-1}g(x)+(x-x_i)^{r_i}g'(x)=(x-x_i)^{r_i-1}P(x),$$
where $P(x)=r_ig(x)+(x-x_i)g'(x)$. Then $x_i$ actually is a root of $f'$; moreover, if $x-x_i$ happens to divide the polynomial $P(x)$, we would get that $x-x_i$ divides $g(x)$, but this contradicts the fact of $r_i$ being the multiplicity of the root $r_i$ of the polynomial $f$.
We found then the roots $\xi_i, \ldots, \xi_{m-1}$ and $x_1, \ldots, x_l$ of the polynomial $f'$. Since
$$(m-1)+(r_1-1)+ \cdots +(r_l-1)=(r_1+ \cdots +r_l)+(m-l)-1= \delta(f)-1=\delta(f'),$$
we conclude that
$f'(x)=(x-\xi_1)\cdots(x-\xi_{m-1})(x-x_1)^{r_1-1}\cdots(x-x_l)^{r_l-1}$
and we see that all the roots of $f'$ are real numbers.
A: A different explanation.
It is an immediate consequence of the Gauss-Lucas theorem stating that the roots of $f'$ are situated in the convex hull of the roots of polynomial $f$, which is a line segment of the real axis.
