General derivative of a polynomial Is the following correct?
$$\frac{d^n}{dx^n} \sum_{k=0}^m a_k x^k = \sum_{k=0}^m \frac{\Gamma(k+1)}{\Gamma(k-n+1)} a_k x^{k-n} \quad n \in \mathbb{N}$$
Obviously this encounters $\frac{1}{\Gamma(1-n)}$ which should be zero for the formula to work and indeed
$$\lim \limits_{x \to n} \frac{1}{\Gamma (1-n)} = 0$$
but is this enough to make the formula rigorous or are there other flaws?
 A: The gamma function has poles at zero and the negative integers and $\frac{1}{\Gamma(x)}$ is an entire function with zeros at these points. This makes the representation below valid.
\begin{align*}
\frac{d^n}{dx^n} \sum_{k=0}^m a_k x^k = \sum_{k=0}^m \frac{\Gamma(k+1)}{\Gamma(k-n+1)} a_k x^{k-n} \quad m, n \in \mathbb{N}
\end{align*}

Nevertheless we can do the same job with simpler means, since we have for $m,n$ non-negative integers:
\begin{align*}
\frac{d^n}{dx^n} \sum_{k=0}^m a_k x^k&=\sum_{k=0}^ma_k\frac{d^n}{dx^n}x^k\\
&=\sum_{k=0}^ma_k \color{blue}{k^{\underline{n}}}x^{k-n}
\end{align*}
where $k^{\underline{n}}=k(k-1)\cdots (k-n+1)$ denotes the falling factorial.

A: Your formula is correct. If you only want to use the classic factorial function, you may write the formula is
$$\frac{\mathrm{d}^n}{\mathrm{d}x^n} \sum_{k=0}^m a_k x^k = 
\sum_{k=0}^m \Theta(k-n) a_k \frac{k!}{(k-n)!}x^{k-n}, \quad n \in \mathbb{N},$$
where $\Theta$ is the Heaviside function. It is an (easy) function that returns $0$ or $1$. I modified the input in the formula, so that it gives us the values that we need.
