closed form of $\prod_1^n (x+k)$

Question

Is there a closed form for the product $$ \prod_{k=1}^n (x+k) $$

I'm trying to find a nice formula for $$ \Gamma(z+n) = f(n)\Gamma(z) $$ for some appropriate f.

Answer 1

If $x$ is a non-negative integer $$ \prod_{k=1}^n (x+k) = \frac{(x+n)!}{x!} $$ Else, the product can be written as a polynomial $$ \prod_{k=1}^n (x+k) = C_nx^n + C_{n-1}x^{n-1} + C_{n-2}x^{n-2} +\ ...\ + C_0x^0 $$ The coefficients $C_n$ and $C_0$ are trivial $$C_n = 1$$ $$C_0 = n!$$ The coefficients $C_{n-1}$ and $C_1$ can be given by $$C_{n-1} = \sum_{k=1}^nk = \frac{n(n+1)}{2}$$ $$C_1 = \sum_{k=1}^n\frac{n!}{k} = n!\sum_{k=1}^n\frac{1}{k}$$ The coefficient $C_{n-2}$ is given as $$C_{n-2} = \left(\sum_{k=1}^nk\right)\left(\sum_{k=1}^nk\right) - \left(\sum_{k=1}^nk^2\right) = \left(\frac{n(n+1)}{2}\right)^2 - \left(\frac{n(n+1)(2n+1)}{6}\right) = \frac{n(n+1)(n-1)(3n+2)}{12}$$

The remaining coefficients must be calculated using certain advanced techniques of combinatorics, which I'm not familiar with. My answer is incomplete as of now, but I hope that someone completes my answer in the future.