Clean way to expand product $\prod_{k=1}^n (1 + x_k)$ 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.

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