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TopNewestOne way to explain this is similar to a way to explain why \(x^0=1\) \((x\neq 0)\):

We know that \[n!=n(n-1)!\] \[1!=1(0)!\] \[0!=1\] – Daniel Chiu · 3 years, 3 months ago

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the answer to this problem can be found by the gamma function which is defined as the integration of (e^-x).(x^m-1) where x varies from 0 to infinity; results in gamma m (m>0).. which is also equivalent to (m-1)! putting m=1 and solving the integration we get 0! = 1 ! abhhi bhool jaa engineering mein jayega tab indirectly iska proof mil jayega .. !! :) – Ramesh Goenka · 3 years, 3 months ago

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– Bodhisatwa Nandi · 2 years, 11 months ago

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Ultimately it's all just a convention. When \(0! = 1\), many things are simplified; for example, the binomial formula \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) holds true even for \(k=0,n\), and it follows naturally from the identity \(n! = (n-1)! \cdot n\) for \(n = 1\). You can freely define it otherwise, but things become more complicated then (for example, the binomial formula works "for \(1 \le k \le n-1\) only; if \(k = 0,n\), the result is \(1\)").

The same thing applies for, for example, \(x^0 = 1\) for nonzero \(x\), or that \(1\) is not a prime number, or that \(0\) is even. They are just definitions and you're free to change them, but they generally make things more complicated to state if their definitions are changed. – Ivan Koswara · 3 years, 3 months ago

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