Signed Stirling numbers of the first kind

We define the following

$$s(n,k) = (-1)^{n-k} \left[n\atop k\right]$$

Prove the following

$$(x)_n = x(x-1)(x-2)\cdots (x-n+1) = \sum_{k=0}^n s(n,k)x^k$$

$$\textit{proof}$$

We already proved that

$$x^{(n)} = \sum_{k=0}^n \left[n\atop k\right] x^k$$

Which can be rewritten as

$$(-1)^n (-x)^{(n)} = \sum_{k=0}^n \left[n\atop k\right] (-1)^{n-k}x^k$$

Now use that

$$(-1)^n (-x)^{(n)} = (x)_n$$

and

$$s(n,k) = (-1)^{n-k} \left[n\atop k\right]$$

Which implies

$$(x)_n = \sum_{k=0}^n s(n,k)x^k$$

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