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Let $\alpha\in \mathbb{Z}_p$ be an $p$-adic integer and define for $n\in \mathbb{Z}_{\geq 0}$ ${\alpha\choose n} := \frac{\alpha(\alpha-1)\cdot\ldots\cdot(\alpha-n+1)}{n!}.$

This is again a $p$-adic integer, and we can define $\alpha_1:=\sum_{n=0}^\infty \overline{{\alpha \choose n}} p^n$ and $\alpha_2:=\sum_{n=0}^\infty \overline{{\alpha \choose p^n}} p^n,$

where $\overline{(\cdot)}$ means reduction modulo $p$. How are $\alpha,\alpha_1,\alpha_2$ related? Are there formulas expressing this relation?

Thanks a lot!

Edit: Some thinking led me to the following conclusion: Using continuity of $x\mapsto {x\choose p^n}$ as a function $\mathbb{Z}_p\rightarrow \mathbb{Q}_p$ and Lucas' Theorem it follows that $\alpha=\alpha_2$. Does this seem correct?

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    You can determine whether $p$ divides $\binom{m+n}{n}$ by seeing whether there are any carries when adding $m$ and $n$ in base $p$. $\binom{\alpha}{p^n}$ is particularly simple, since $p^n$ in base $p$ is very easy.2011-03-02

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