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.
 A: Let us first examine some examples to guess the general case:$$(x-r_1)(x-r_2)=x^2-(r_1+r_2)x+r_1r_2$$ $$(x-r_1)(x-r_2)(x-r_3)=x^3-(r_1+r_2+r_3)x^2+(r_1r_2 + r_1r_3+r_2r_3)x-r_1r_2r_3.$$So, we can guess the following identity:$$\prod_{i=1}^n(x-r_i)=\sum_{k=0}^n\sum_{1 \le j_1 \lt ... \lt j_k \le n}(-1)^kr_{j_1} ... r_{j_k}x^{n-k}.$$Let us prove the claim by induction.
The base case is trivial. So, let us assume that the claim is correct for $n=m$, that is,$$\prod_{i=1}^m(x-r_i)=\sum_{k=0}^m\sum_{1 \le j_1 \lt ... \lt j_k \le m}(-1)^kr_{j_1} ... r_{j_k}x^{m-k}.$$ So, we need to prove the claim for $n=m+1$ as follows.$$\prod_{i=1}^{m+1}(x-r_i)=\left ( \prod_{i=1}^m(x-r_i) \right ) \left ( \vphantom{\prod_{i=}^n} x-r_{m+1} \right )$$ $$=\left (\sum_{k=0}^m\sum_{1 \le j_1 \lt ... \lt j_k \le m}(-1)^kr_{j_1} ... r_{j_k}x^{m-k} \right )  \left ( \vphantom{\prod_{i=}^n} x-r_{m+1} \right )$$ $${=\sum_{k=0}^m\sum_{1 \le j_1 \lt ... \lt j_k \le m}(-1)^kr_{j_1} ... r_{j_k}x^{(m+1)-k} -\sum_{k=0}^m\sum_{1 \le j_1 \lt ... \lt j_k \le m}(-1)^kr_{j_1} ... r_{j_k}r_{m+1}x^{m-k}}$$ $$=\left (x^{m+1}+\sum_{k=1}^m\sum_{1 \le j_1 \lt ... \lt j_k \le m}(-1)^kr_{j_1} ... r_{j_k}x^{(m+1)-k} \right ) - \left ( \sum_{k=0}^{m-1}\sum_{1 \le j_1 \lt ... \lt j_k \le m}(-1)^kr_{j_1} ... r_{j_k}r_{m+1}x^{m-k}+ (-1)^m r_{j_1} ... r_{j_m}r_{m+1} \right )$$ $$=\left (x^{m+1}+\sum_{k=1}^m\sum_{1 \le j_1 \lt ... \lt j_k \le m}(-1)^kr_{j_1} ... r_{j_k}x^{(m+1)-k} \right ) - \left ( \sum_{k=1}^{m}\sum_{1 \le j_1 \lt ... \lt j_k \le m}(-1)^{k-1}r_{j_1} ... r_{j_{k-1}}r_{m+1}x^{m-(k-1)}+ (-1)^m r_{j_1} ... r_{j_m}r_{m+1} \right )\tag{*}\label{*}$$ $$= \left ( x^{m+1} + (-1)^m r_{j_1} ... r_{j_m}r_{m+1} + \sum_{k=1}^m (-1)^k \left ( \sum_{1 \le j_1 \lt ... \lt j_k \le m}r_{j_1} ... r_{j_k}+ r_{j_1} ... r_{j_{k-1}}r_{m+1} \right ) x^{(m+1)-k} \right )$$ $${= \left ( x^{m+1} + (-1)^m r_{j_1} ... r_{j_m}r_{m+1} + \sum_{k=1}^m (-1)^k  \sum_{1 \le j_1 \lt ... \lt j_k \le m+1}r_{j_1} ... r_{j_k} x^{(m+1)-k} \right )}$$ $$=\sum_{k=0}^{m+1}  \sum_{1 \le j_1 \lt ... \lt j_k \le m+1}(-1)^k r_{j_1} ... r_{j_k} x^{(m+1)-k}.\tag{**}\label{**}$$Thus, by induction we proved that for any natural number $n$ the following identity holds:$$\prod_{i=1}^n(x-r_i)=\sum_{k=0}^n\sum_{1 \le j_1 \lt ... \lt j_k \le n}(-1)^kr_{j_1} ... r_{j_k}x^{n-k}.$$

Footnote
\ref{*} is followed from the following property of summation:$$\sum_{i=m}^nA_i=\sum_{i=m+1}^{n+1}A_{i-1}.$$
\ref{**} is followed from considering the fact that for any fixed $k$ one can decompose the sum $\sum_{1 \le j_1 \lt ... \lt j_k \le m+1} r_{j_1} ... r_{j_k}$ into two sums: (i) the sum of terms not containing $r_{j_{m+1}}$, that is, $\sum_{1 \le j_1 \lt ... \lt j_k \le m} r_{j_1} ... r_{j_k}$, and (ii) the sum of terms containing $r_{j_{m+1}}$, that is $\sum_{1 \le j_1 \lt ... \lt j_k \le m} r_{j_1} ... r_{j_{k-1}}r_{j_{m+1}}$.
A: 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...
