[Edit: "Suppose K is a field complete with respect to a discrete valuation, with valuation ring $\mathcal{O}$."]
I'm trying to solve the following problem: let $L/K$ be a finite extension of fields, and $x \in L$. Show if $N_{L/K}(x) \in \mathcal{O}$ then $N_{L/K}(1+x) \in \mathcal{O}$, where $N_{L/K}$ denotes the norm (i.e. the product of embeddings $\sigma_i:L \to \overline{K}$) and $\mathcal{O}$ denotes $\{x:|x|_K \leq 1\}$ under the absolute value $|\cdot|_K$ on $K$.
So, I did the obvious thing, and wrote out the product
$N_{L/K}(1+x)=\prod_{i=1}^n \sigma_i(1+x)=\prod_{i=1}^n (\sigma_i(1)+\sigma_i(x))=\prod_{i=1}^n (1+\sigma_i(x))$ $= \prod_{i=1}^n \sigma_i(x)+\sum_{j=1}^n \prod_{i\neq j} \sigma_i(x) + \sum_{j=1}^n \sum_{k=1}^n \prod_{i\neq j,k} \sigma_i(x)+\ldots +1$
where $\prod_{i=1}^n \sigma_i(x)=N_{L/K}(x)$. This unfortunately got me nowhere: you can pull a $N_{L/K}(x)$ out of every term to write the sum in a slightly different way, but I can't see any way to show that this should also be in $\mathcal{O}$. Obviously the norm is multiplicative but pulling out this factor with norm at most 1 didn't help anything. Could anyone help? I expect I'm probably missing something painfully obvious. Thanks! -BTS