Sort of Binomial Expansion I was trying to find a general formula for expanding the product: $$\prod_{i=1}^k (a+ib)$$ where $a, b \in \mathbb{R}$. The first few expansions are as follows: $$\prod_{i=1}^1 (a+ib) = a + b$$ $$\prod_{i=1}^2 (a+ib) = a^2+3ab+2b^2$$ $$\prod_{i=1}^3 (a+ib) = a^3 +6a^2b +11ab^2 +6b^3$$ At the time, I thought that there may be a connection between this product and the binomial expansion. I still think I am correct in thinking this. However, my combinatorial skills are not up to par. Anyways, I found a recursive formula that can generate the coefficients of the expansion. To steal from classical binomial coefficient notation, let  $\binom{0}{0} = 1$, $\binom{1}{0} = 1$, $\binom{1}{1} = 1$, and $\binom{k}{-1} = 0$, with $k\ge0$ be the initial values and have  $$\binom{n}{k} = n\binom{n-1}{k-1} + \binom{n-1}{k}$$ be the definition of this symbol during this question. Notice that this is very similar (but not identical) to the recursive definition of the original binomial expansion, which leads to all sorts of great mathematical fun. With this new knowledge and notation, I easily got the expansion. It is $$\prod_{i=1}^k (a+ib) =\sum_{i=0}^k \binom{k}{i}a^{k-i}b^i.$$ This is a pretty formula but I would feel better if I knew more than just the recursive definition. Help!!! So the question is this: How can I obtain a non-recursive definition from this?
I found out how cool this recursive definition is by replacing the traditional binomial coefficients with their corresponding "pseudo-binomial coefficients" (bad name, I know) on Pascal's Triangle.
I sort of feel bad for not proving any of the things I said here, but hopefully you will forgive me after you search for all the pretty patterns that come up when you inspect this new sort of triangle! I also doubt that I am the first one to think of this, so it would be great if someone had information on this stuff. (I have not found any information on it, but I didn't try very hard sooo...)Thank you so much in advanced for your patience and your diligence!
 A: Notice that:
$$\begin{align}\prod_{i=1}^{k}(a+i b) & = b^k\prod_{i=1}^{k}(\tfrac a b + i) \\ & = b^k(\tfrac a b +1)(\tfrac a b +2)\cdots(\tfrac a b + k)\end{align}$$
Now if $\tfrac ab$ were an integer, say $n$, then we would have: $$b^k\prod_{i=1}^k(n+i)= \frac{(n+k)!\,b^k}{n!}$$
Now the factorial function can be extended to real (and complex) arguments, through the complete gamma function.
$$\Gamma(z) = \int_0^\infty e^{-t} t^{z-1}\operatorname{d}t$$
And for natural number arguments: $\Gamma(n+1)=n!$
So what we have is:$$\prod_{i=1}^k (a+i b) = \frac{b^k \Gamma(\tfrac ab + k+1)}{\Gamma(\tfrac ab+1)}$$
This ratio occurs often enough to be given a symbol: Pochhammer, and it's summation expansion is:
$$(x)_k=\frac{\Gamma(x+k)}{\Gamma(x)}=\sum\limits_{i=1}^k (-1)^{k-i}\operatorname{s}(k,i)x^i$$
Where $\operatorname{s}(k,i)$ is a Stirrling number of the first kind, which is defined by the recurrence relation:
 $$\begin{align}
\operatorname{s}(k+1,i) & = \operatorname{s}(k,i-1) - k \operatorname{s}(k,i), &\text{ for }1\leq i\leq k 
\\ \operatorname{s}(k,i)&=\sum_{j=i}^k k^{j-i} \operatorname{s}(k+1,i+1),&\text{ for } i\geq 1
\\ \binom{i}{r}\operatorname{s}(k,i)&=\sum_{j=i-r}^{n-r}\binom{k}{j}\operatorname{s}(k-j,r)\operatorname{s}(j,i-r),&\text{ for }0\leq r \leq i
\end{align}$$ 
And... well, that should give you a start for further research topics.
