Obtaining an expression between $s'(n,r)$ and $s(n,r)$

Question

I've a doubt in this:

We're given $[x]_n=(x)(x-1)\ldots (x-(n-1))$ and $[x]^n=(x)(x+1)\ldots (x+n-1)$ .
Now as we can write :
$[x]_n=(x)(x-1)\ldots (x-(n-1))=a_0+a_1x+a_2x^2\ldots +a_nx^n$.

Here the coefficient of $x^r$ in expansion i.e. $a_r$is denoted as $s(n,r)$.

Hence we write:
$$[x]_n=\sum_{r=0}^{n}s(n,r)x^r$$ Similarly my notes state
$$[x]^n=\sum_{r=0}^{n}s'(n,r)x^r$$

then the following relation is directly written in my notes: $$s'(n,r)=(-1)^{n-r}s(n,r)$$

How can we obtain this last relation is what I can't understand....Please help how to obtain this expression...

Answer 1

Hint. You have $$ \begin{align}[x]_n&=(x)(x-1)\ldots (x-(n-1))\\ &=(-1)^{n}(-x)((-x)+1)\ldots ((-x)+n-1)\\ &=(-1)^{n}[-x]^n \end{align}$$ then you identify your polynomials.

Hope you can take it from here.