Consider the good old Pascal’s triangle:

pascal

Except for the first row, take the alternating sum of the entries. So for the second row we have $ 1 – 1 = 0$. For the third row we have $ 1 – 3 + 3 – 1 = 0$. For the fourth row we have $ 1 -4 + 6 – 4 + 1 = 0$, etc. So it seems that we have the following for $ n > 0$:

$ \sum_{i=0}^n (-1)^n\binom{n}{i} = 0$

One can use the binomial theorem to prove this. In the process of reviewing some material on regular sequences, I came up with a slightly different proof that could be the most pretentious and ridiculous proof of this fact. (However, even more ridiculous proofs using heavy machinery would be welcome in the comments.) The reader may wish to consult the previous post describing the Koszul resolution before reading onwards.
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