prove $S(n,1) = -1^{n-1}(n-1)!$ (Stirling numbers of the first kind)

Question

what is have managed so far using this formula:


$S(n + 1,k + 1) = \sum_{ν=k}^{n} -1^{n-ν}F_{n-ν}(n)S(ν,k) \forall k=0,1,...,n $
$n\epsilon \mathbb{N}$
and $F_{n}(x) = x(x-1)...(x-n+1)$
$n\epsilon \mathbb{N^{*}}$


$S(n,1) =\sum_{ν=0}^{n-1} -1^{n-1-0}F_{n-1-0}(n-1)S(ν,k)$

since $S(n,n) = 1$ and ν = κ $S(ν,k) = 1 $
So we have that:

$\sum_{ν=0}^{n-1} -1^{n-1-0}F_{n-1-0}(n-1)S(ν,k) = \sum_{ν=0}^{n-1} -1^{n-1}F_{n-1}(n-1) =$
$\sum_{ν=0}^{n-1} -1^{n-1}(n-1)(n-2)...(n-1-n+1+1)=\sum_{ν=0}^{n-1} -1^{n-1}(n-1)!$

Inside the sum is what i want to get as a final asnwer.Any ideas on how to get it without a sum?Did i use a wrong formula?An explanation to every answer would be appreciated thanks in advance!