Lets say I have a sum that states the following $$ \sum_{j=0}^{k-c} {k-c \choose j}\ln(a)^{k-c-j} \frac{d^j}{dx^j}[(x)_c] $$ where $(x)_c$ is the falling factorial such that
$$ (x)_c = x(x-1)(x-2)\cdots(x-c+1) $$
How can I evaluate this summation?
Viewing the summations partially, I know that
$$ \sum_{j=0}^{k-c}{k-c \choose j}\ln(a)^{k-c-j} = \left(\frac{1}{\ln(a)}+1\right)^{k-c} \ln(a)^{k-c} $$ But it is possible that $k-c > c$ and therefore having the summation above be incomplete due to differentiating a constant, resulting in an incomplete gamma function result seen here: http://www.wolframalpha.com/input/?i=sum+%28%28k-c%29%21%2F%28%28k-c-r%29%21%29%29ln%28y%29%5E%28k-c-r%29%2C+r%3D0..k-c%2Bn
An interesting thing to note is that taking multiple derivatives of the falling factorial x results in the sum of the permutations that the falling factorial can be presented in, multiplied by the factorial of the amount of times differentiated. an example would be: $$ \frac{d}{dx} [x(x-1)(x-2)(x-3)] = 1![x(x-1)(x-2)+x(x-1)(x-3)+x(x-2)(x-3)+(x-1)(x-2)(x-3)] $$
$$ \frac{d^2}{dx^2} [x(x-1)(x-2)(x-3)] = 2![x(x-1)+x(x-2)+x(x-3)+(x-1)(x-2)+(x-1)(x-3)+(x-2)(x-3)] $$ Note that the number of terms produced by taking $n$ derivatives of $(x)_c$ is directly proportional to ${n \choose c}$
Is there no defined way to express this summation?
Thank you for any information in advance.