<s>[INST] Coefficient of $x^{n-2}$ in $(x-1)(x-2)(x-3)\dotsm(x-n)$ Question

Find the coefficient of $x^{n-2}$  in the expression $$(x-1)(x-2)(x-3)\dots(x-n)~~.$$

My approach

The coefficient of $x^n$ is $1$. The coefficient of  $x^{n-1}$ is $- \frac{n(n+1)}{2}$

But I cannot proceed from here.
I would appreciate any help.
 [/INST] For $~x^{n-1}~$ the coefficient would be sum of the roots that is $~1+2+3+4+5+\cdots+n~$ for $~x^{n-2}~$ coefficient would be sum of roots taken two at a time that is $~1.2+2.3+3.4+4.5+\cdots+(n-1)n~$ so let it be $~S~$ them $~S~$ can be find using steps below
$$(1+2+3+4+5+\cdots+n)^2=(1^2+2^2+3^3+\cdots+n^2)+2(1.2+2.3+3.4+\cdots+(n-1)n)$$ 
So 
$$(1+2+3+4+5+\cdots+n)^2=(1^2+2^2+3^3+\cdots+n^2)+2(S)$$ 
So this can be find as we know sum of $~n~$ natural no 
That is $~\frac{1}{2}\{n(n+1)\}~$ and sum of squares of first $~n~$ natural number is $~\frac{1}{6}\{n(n+1)(2n+1)\}~$ so this value can be found
Thanks for opportunity to solve the problem
</s>