3 ms·
0! really is 1 in a much more reasonable sense. The empty product is the multiplicative identity. The continuous notion of ! also agrees: http://en.wikipedia.or
by anonymoushn 13y ago
0! really is 1 in a much more reasonable sense. The empty product is the multiplicative identity. The continuous notion of ! also agrees: http://en.wikipedia.org/wiki/Gamma_function http://en.wikipedia.org/wiki/Gamma_function
- gordaco 13y agoAnd it's far easier to prove: 1!=1, (n+1)!=n!·(n+1), thus 0!=1!/1=1.