Find, in the simplest form, the coefficient of x^n in the binomial expansion of (1-x)^(-6).

Hence, find the coefficient of x^6 and x^7 in (1+2x+3x^2+4x^3+5x^4+6x^5+7x^6)^3

Attempt:

The generalised binomial theorem,

$\displaystyle \sum^{\infty}_{k=0}\frac{-6(-6-1)(-6-2)...(-6-r+1)}{r!}(1)^{-6-r}(-x)^r$

So when r=n, the coefficient of x^n will be $\displaystyle -\frac{-6(-6-1)(-6-2)...(-6-r+1)}{r!}$

Can that be further simplified?

And how can i go about in the second part?