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Jun 12, 2014 at 2:19 comment added Tom Goodwillie Well, you can start the induction with the trivial $n=0$ case.
Jun 12, 2014 at 1:56 comment added KConrad Concerning the comment at the end about proofs by induction, the case $n=1$ is not always trivial, but sometimes is in fact where all the meat of the argument lies. Consider the proof that if $A$ is a UFD then $A[X_1,\dots,X_n]$ is a UFD. Only the case $n=1$ is hard.
Sep 11, 2011 at 18:35 history edited roy smith CC BY-SA 3.0
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Sep 11, 2011 at 18:23 history answered roy smith CC BY-SA 3.0