Expanding linear factors of a polynomial Expanding
$$a(x-r_1)(x-r_2)\cdots (x-r_n)$$
should give
$$ax^n-a(r_1+r_2+\cdots r_n)x^{n-1}+a(r_1 r_2+r_1 r_3+\cdots r_{n-1}r_n)x^{n-2}+\cdots (-1)^{n}ar_1 r_2\cdots r_n$$
but I fail to prove it. I was only able to do cases $n=1$, $n=2$ and $n=3$ (and even the result for $n=3$ seems a bit different):
$$\begin{align*}a(x-r_1)&=ax-ar_1\\
a(x-r_1)(x-r_2)&=(ax-ar_1)(x-r_2)\\&=ax^2-axr_2-axr_1+axr_1 r_2\\&=ax^2-a(r_1+r_2)x+ar_1 r_2 x\\a(x-r_1)(x-r_2)(x-r_3)&=(ax^2-a(r_1 +r_2)x+ar_1 r_2 x)(x-r_3)\\&=ax^3-a(r_1 +r_2)x^2+ar_1 r_2 x^2-ar_3 x^2+ar_3 (r_1 +r_2)x-ar_1 r_2 r_3x\\&=ax^3-x^2(a(r_1 +r_2)-ar_1 r_2+ar_3)+x(ar_3 (r_1+r_2)-ar_1 r_2 r_3)\\&=ax^3-a(r_1+r_2-r_1r_2+r_3)x^2+a(r_1r_3+r_2r_3-r_1r_2r_3)x\end{align*}$$
Could someone help me to prove the general case for all $n$?
Edit: There's an error in my computation.

Let's drop the $a$ as it is common everywhere
Now, without expanding, let's analyze:
How many ways to make an $x^n$ term? Only 1 way (taking $x$ from each bracket) 
How many ways to make an $x^{n-1}$ term? You can take $n-1 $ powers of $x$ and one other term from any bracket . Thus the coefficient is $-r_1-r_2...$
Continue as above...
