Finding coefficient of $x^8$ There is a question I am having difficult time with. Please help me to find the coefficient of $x^8$ in the polynomial $$(x-1)(x-2)(x-3)........(x-10)$$ I don't even know to start but I think it's related to binomial theorem but again there are no powers here. Any help is appreciated. Thank you.
 A: The coefficient of $x^8$ of $\prod_{k=1}^{10}(x-k)$ is given by
$$\sum_{1\leq j<k\leq 10}kj=\frac{1}{2}\left(\left(\sum_{k=1}^{10}k\right)^2-\sum_{k=1}^{10}k^2\right)=
\frac{1}{2}\left(55^2-385\right)=1320.$$
See also Vieta's formulas.
In other words, here the coefficient of $x^8$ is given by the sum of all products of two distinct numbers in $\{1,2,..,10\}$. This sum can be obtained by squaring $(1+2+3+\dots+10)$ and then by throwing away all the squares $1,4,\dots, 100$. The result should be divided by 2 (they are double products).
A: First, we see how we get the powers of $x^8$ thru the traditional method of "expanding out".
To get powers of $x^8$, we will actually multiply 8 $x$'s together with two numbers, e.g. 
$x\dots x\cdot (-2)(-3)=6x^8$
So this boils down to finding all the sums of products of two distinct numbers, i.e. $1\times 2+1\times3+\dots+9\times 10$.
First we find out what is $1(2)+1(3)+1(4)+\dots+1(10)$. Surely, it is 54. One other way to write it is $(1+2+\dots+10)-1=54$
Next, we find the product of two distinct numbers, containing 2, i.e.: 
$2(2)+2(3)+2(4)+\dots+2(10)$. This is $2(1+3+\dots+10)=2(1+2+\dots+10)-2^2$.
The shortcut to this is to observe that the sum is actually:
$(1+2+\dots+10)(1+2+\dots+10)-(1^2+2^2+\dots+10^2)=55(55)-385=2640$
Note that we have double counted: for instance $1\times 2$ and $2\times 1$ are counted twice.
So dividing by 2 gives 1320 as desired.
