If $ \scriptstyle p $ is a prime and $ \scriptstyle p\ \mid\ a_1a_2\cdots a_n $, then $ \scriptstyle p\ \mid\ a_1 $ or $ \scriptstyle p\ \mid\ a_2a_3\cdots a_n $. If $ \scriptstyle p\ \mid\ a_1 $, we are done. Otherwise, $ \scriptstyle p\ \mid\ a_2 $ or $ \scriptstyle p\ \mid\ a_3a_4\cdots a_n $. This continues iteratively until $ \scriptstyle p\ \mid\ a_{n-1}a_n $. Then $ \scriptstyle p\ \mid\ a_{n-1} $ or $ \scriptstyle p\ \mid\ a_n $, and it is clear that $ \scriptstyle p\ \mid\ a_i $ for some $ \scriptstyle1\ \le\ i\ \le\ n $. $ \scriptstyle\Box $
- --Nick Rupley 00:27, 22 January 2009 (UTC)
