Simplifying the product of multiple binomial expansions I have a tricky product I'm trying to expand out into a summation, and I'm not sure how to go forward.
I have two sets of numbers, each containing $n$ elements total. The first set, $\{x_k\}$ are real and positive:
$$ x_k \in \mathbb{R}, \; x_k > 0 \; \forall \; k$$
The second set, $\{y_k\}$, are all naturals:
$$ y_k \in \mathbb{N} \; \forall \; k$$
The expression I'm trying to expand out is a product of different binomial expansions:
$$ f(c) = \prod_{k=1}^{n} (c x_k +1)^{y_k} = \prod_{k=1}^{n} \left( \sum_{m=0}^{y_k} \binom{y_k}{m}c^{m} x_{k}^{m} \right)$$
where $c>0$.
When fully expanded, we can only group by powers of $c$, which must run between zero and $\sum_{k}^{n} y_k$. Letting
$$ Y = \sum_{k=1}^{n} y_k $$
we should be able to express $f(c)$ as
$$ f(c) = \sum_{m=0}^{Y} \alpha_m c^{\beta_m}$$
Now I'm stuck for how to calculate the coefficients $\alpha_m$ and exponents $\beta_m$.
Anyone have any advice?
Thanks!
 A: Using the same index of summation $m$ in all the sums might be a bit confusing, as these different indices could all take different values in the product.  It therefore might be better to put subscripts on those indices, so we would have
$$ f(c) = \prod_{k=1}^n \left( \sum_{m_k = 0}^{y_k} \binom{y_k}{m_k} c^{m_k} x_k^{m_k} \right). $$
When we exchange the summation and the product, the sum will now be over all possible choices of the indices $m_k$.  Hence it will be a sum over the vectors $\vec{m} = (m_1, m_2, ..., m_k)$, where each $m_k$ satisfies $0 \le m_k \le y_k$.  Hence we have
$$ f(c) = \sum_{\vec{m} \in \Lambda} \prod_{k=1}^n \left( \binom{y_k}{m_k} c^{m_k} x_k^{m_k} \right), $$
where $\Lambda = \mathbb{Z}^n \cap \prod_{k=1}^n [0, y_k]$.
We can now collect the factors of $c$, and we would get
$$ f(c) = \sum_{\vec{m} \in \Lambda} c^{\sum_{k=1}^n m_k} \prod_{k=1}^n \left( \binom{y_k}{m_k} x_k^{m_k} \right). $$
Now different vectors $\vec{m} \in \Lambda$ could result in the same power $p = \sum_{k=1}^n m_k$ of $c$.  If you would like, you could group them together; let $\Lambda_p = \left\{ \vec{m} \in \Lambda: \sum_{k=1}^n m_k = p \right\}$.  As you pointed out, the possible powers are the integers between $0$ and $Y = \sum_{k=1}^n y_k$.  We would then have
$$ f(c) = \sum_{p = 0}^Y \alpha_p c^p, $$
where the coefficients $\alpha_p$ are given by collecting the terms from $\Lambda_p$:
$$ \alpha_p = \sum_{\vec{m} \in \Lambda_p} \prod_{k=1}^n \binom{y_k}{m_k} x_k^{m_k}. $$
