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Is there a clean way to write the expansion of:

$$\prod_{k=1}^n (1 + x_k) = (1 + x_1)(1 + x_2)\dots(1 + x_n)$$

The expansion may be written:

$$1+\sum_{1\le i \le n} x_i +\sum_{i \le n}\sum_{j\lt i} x_j x_i + \sum_{i \le n}\sum_{j\lt i}\sum_{k\lt j} x_kx_jx_i+\cdots+ x_1x_2\cdots x_n$$

But it would be nice if there were a more compact way of writing this. I tried making use of the Levi-Civita tensor over an indexed set:

$$1 + \sum_{k=1}^n \sum_{(i_j)_{j \in \{1 \dots k\}} \in \{1 \dots n\}} \frac{1}{2} |\epsilon_{i_1 \dots i_k}| \prod_{j=1}^k x_{i_j}$$

But that seems a little messy.

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As explained in Richard Stanley book we have: $$\prod_{k=1}^n (1 + x_k) = (1 + x_1)(1 + x_2)\dots(1 + x_n) =\sum_{A\subseteq [n]}\prod_{i\in A}x_i$$

where $\displaystyle{\prod_{i\in\varnothing}x_{i}:=1}$ as Ethan remarks.

More generally, you can consider even the multiset case:

$$\prod_{k=1}^n (1 + x_k+x_k^{2}+\cdots) = (1 + x_1+x_1^{2}+\cdots)(1 + x_2+x_2^{2}+\cdots)\dots(1 + x_n+x_n^{2}+\cdots)$$ $$=\sum_{(M,\,\mu)\subseteq [n]}\prod_{i\in (M,\mu)}x_i^{\mu(i)}$$

where $(M,\mu)$ is a multi-subset of $[n]=\{1,2,\ldots,n\}.$

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I'd probably use elementary symmetric polynomials. However, here is an equivalent representation more generally for any (possibly non-commutative) semi-ring equipped with $+,\times$ and an identity $I$ Namely if we denote the family of all order perserving injections from $\{1,2,\ldots, k\}$ to $\{1,2,\ldots, n\}$ by the set $[n,k]$ so that we get:

$$\varphi\in [n,k]\iff \left[\varphi:\{1,\ldots, k\}\to \{1,\ldots, n\} \text{ is injective}\right]\land \left[k<n\implies \varphi(k)<\varphi(k+1)\right]$$

Then we can express the following product explicitly:

$$\prod^n_{k=1}(I+a_k)=I+\sum_{k=1}^n\left(\sum_{\varphi\in[n,k]}\prod_{j=1}^ka_{\varphi(j)}\right)$$

Noting that $\left\{\text{rng}(\varphi):\varphi\in [n,k]\right\}$ is the set of all $k$ element subsets of $n$ therefore $|[n,k]|=\binom{n}{k}$.


If multiplication is commutative, the indices reduce to subsets of $\{1,2,3,\ldots, n\}$ so we can write:

$$\prod_{k=1}^n(I+a_k)=I+\sum_{\substack{S\subseteq \{1,2,\ldots, n\}\\ S\neq \emptyset}}\prod_{a\in S}a$$

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