If $\alpha $ is a double root of the following equation then how to tackle it? I am stuck with the following problem that says: 

If $\alpha $ is a double root of the equation $x^n+p_1x^{n-1}+........+p_n=0,\,\,\,\,$prove that $\alpha $ is also a root of the equation $p_1x^{n-1}+2p_2x^{n-2}+3p_3x^{n-3}+...........+np_n=0$.

My try: Since  $\alpha $ is a double root of the equation $x^n+p_1x^{n-1}+........+p_n=0,\,\,\,\,$ $f(\alpha)=0,\,\, f'(\alpha)=0$.
Now, I don't know how to progress to get the desired result. Please help.
 A: Hint: You seem to have realized that the combination $f(x) - xf'(x)$ has $\alpha$ as a root. And it's close to the solution, but it does not quite give the polynomial you're asked about.
For instance, the coefficient of $x^n$ in $xf'(x)$ is $n$, and in whatever combination you ultimately choose, you want that term to go away. How can you manipulate $f(x)$ to make the $x^n$ term go away in the combination?
Along a similar vein, you want whatever combination you end up with to have constant term $np_n$, and $xf'(x)$ has $0$ as constant term. How can you manipulate $f(x)$ to make the constant term of the combination into $np_n$?
Think about these two questions. Think of one thing that answers both of them simultaneously. That will most likely solve your problem.
A: Hint 1:

 Can you "shift" it so that the derivative comes out as $(n+k)x^{n+k-1}+(n+k-1)p_1x^{n+k-2}+\dots+kp_nx^{k-1}$?

Hint 2:

 For what value of $k$ does the first coefficient come out as $0$, and the last coefficient come out almost as $np_n$?

