I have asked this question and have come up with a possible answer $$ \frac{d^j}{dx^j}[\frac{(x)_c}{j!}] = e_{c-j}(x,x-1, \cdots ,x-c+1) $$ My first question is, how can I prove this? It seems trivial but I do not know the right approach to doing so. Also can I use the elementary symmetric polynomial to express multiple derivatives? I feel like I can because, well I wouldn't see why I couldn't. I have yet to really find a definitive answer to wether or not I can use terms instead of numbers with elementary symmetric polynomials. This may just be a stupid question I am not sure, and I do apologize if it is.
So if this is true, I can then conclude that:
$$ \frac{d}{dx}\frac{d^j}{df(x)^j}[\frac{(f(x))_c}{j!}] = \frac{d}{dx}[e_{c-j}(f(x),f(x)-1,\cdots,f(x)-c+1)] = (j+1)f'(x) e_{c-j-1}(f(x),f(x)-1,\cdots,f(x)-c+1) $$
If you differentiate $(x)_c$ $j$ times, using the product rule, what you will get is a sum over all the different ways to choose $j$ of the $c$ terms, of the product of the remaining terms. This product is exactly $e_{c-j}(x,x-1,\dots,x-c+1)$, and that product is obtained $j!$ times because you have $j!$ different orders in which you could choose the $j$ terms to remove.