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Let $A$ be a local ring with maximal ideal $m$ that is $m$-adically complete, and assume $1/2 \in A^\times$. I've read in several places that for any $x \in m$, a square root of $1 + x$ in $A$ is given by the binomial series

$ \sum_{n = 0}^\infty {{1/2} \choose n} x^n, $ where ${{1/2} \choose n} = \frac{(1/2)(1/2 - 1) \cdots (1/2 - n + 1)}{n!}$. I don't understand why these binomial coefficients make sense in $A$. We have $n!$ in the denominator, so why is $n! \in A^\times$?

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    Julian: You could post my comment as an answer if you wish.2013-06-25

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As KCd remarks:

A ratio being in $A$ does not require that the denominator has to be a unit. For example, $15/5$ is an integer even though $5$ is not a unit in the integers. Or, to give an example with a local ring, $15/5$ is in $\Bbb Z_5$ (the $5$-adic integers) even though $5$ is not a unit in $\Bbb Z_5$. The point is that the rational number $1/2 \choose n$ in reduced form has a denominator that is a power of $2$, so it makes sense in $A$. See this question for details.