I spend 4 hours trying to solve it. I believe that I am either stuck or I am approaching the problem at the wrong angle. Here is my challenge:
$$\frac{\partial^m }{\partial x^m}\left [ x^{-a}\gamma \left ( a,x \right ) \right ]=\left ( -1 \right )^{m}x^{-a-m}\gamma\left ( a+m,x \right )$$ This is from Arfken Textbook.
I used the following series that represents the incomplete gamma function
$$\gamma \left ( a,x \right )=x^{a}\sum_{n=o}^{\infty} (-1)^{n}\frac{x^{n}}{n!(a+n)}$$
Then I proceeded as follows:
$$\gamma \left ( a,x \right )=x^{a}\sum_{n=0}^{\infty} (-1)^{n}\frac{x^{n}}{n!(a+n)}$$ $$x^{-a}\gamma \left ( a,x\right )=\sum_{n=0}^{\infty} (-1)^{n}\frac{x^{n}}{n!(a+n)}$$ $$\frac{\partial^m }{\partial x^m}\left [ x^{-a}\gamma \left ( a,x \right ) \right ]=\frac{\partial^m }{\partial x^m}\sum_{n=0}^{\infty} (-1)^{n}\frac{x^{n}}{n!(a+n)}=\sum_{n=0}^{\infty} (-1)^{n}\frac{x^{n-m}}{n!(a+n)}(n)(n-1)(n-2)(n-3).........(n-m)$$ $$=x^{-a-m}x^{a+m}x^{-m}\sum_{n=0}^{\infty} (-1)^{n}\frac{x^{n}}{n!(a+n)}\frac{(a+m+n)}{a+m+n}n_{(m)}$$ $$=\left ( -1 \right )^{m}x^{-a-m}\gamma\left ( a+m \right )x^{-m}\frac{\left ( a+m+n \right )}{(a+n)}(-n)^{(m)}$$ As you see, I am almost there except that I ended with extra terms that I do not see how to make them cancel out. I believe that my approach is incorrect unless I am missing something. What am I missing? Thanks for assistance.
After two small corrections in your equation, the simplification becomes easy :