We know that "n!=n(n-1)\\ldots1\\;\\;\\forall n\\in\\mathbb{N}". Now, we can infer from this definition that "\\frac{(n+1)!}{n!}=(n+1)". By the finiteness of the Gamma function, we know that "0!" exists. Thus, we can use the above formula; "\\frac{1!}{0!}=1\\implies0!=1"
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