I would like to make
$$ \frac{d^n}{dx^n}[(x)_c] = n! \times e_{c-n}(x,x-1,\cdots,x-c+1) $$
Where $e_{c-n}(x,x-1,x-2,\cdots,x-c+1)$ is the elementary symmetric polynomial function
But lets say that i have
$$ \sum_{k=0}^{n-c}f(k) \frac{d^k}{dx^k}[(x)_c] $$
If $k>c$ then the derivative is zero, But if i replace $\frac{d^k}{dx^k}[(x)_c]$with $k!\times e_{c-k}(x,x-1,\cdots,x-c+1)$ then i get
$$ \sum_{k=0}^{n-c}k!f(k) e_{c-k}(x,x-1,\cdots,x-c+1) $$ If $k>c$ then would $e_{c-k}(x,x-1,\cdots,x-c+1)=0$? If not how can i fix this problem?