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Question Number 25237 by Mr eaay last updated on 06/Dec/17

$$prooovethat0!=1$$

Commented by prakash jain last updated on 07/Dec/17

$$\mathrm{It}\mathrm{is}\mathrm{a}\mathrm{notation}\mathrm{simplication}$$$$\mathrm{that}\mathrm{defines}\mathrm{empty}\mathrm{products}$$$$\mathrm{as}1.$$$$a_1\cdot a_2\cdot a_3.\dots a_n=\underset{i=1}{\prod ^n}{P}_n$$$$P_n=n\cdot P_{n-1}$$$$\mathrm{The}\mathrm{above}\mathrm{notation}\mathrm{we}\mathrm{define}{P}_0=1.$$$$x^n=x\cdot x\cdot \dots x\left(ntimes\right)$$$$x^n=x\cdot x^{n-1}$$$$x^0\mathrm{is}\mathrm{an}\mathrm{empty}\mathrm{product}=1$$$$x^0\mathrm{can}\mathrm{also}\mathrm{be}\mathrm{seen}\mathrm{as}\frac{x^n}{x^n}=x^{n-n}=x^0=1$$$$n!=n\cdot (n-1)!$$$$1!=1.0!$$$$\mathrm{to}\mathrm{have}\mathrm{this}\mathrm{relation}\mathrm{true}0!=1.$$$$\mathrm{All}\mathrm{empty}\mathrm{products}\mathrm{evalute}\mathrm{to}1.0!$$$$\mathrm{is}\mathrm{an}\mathrm{empty}\mathrm{product}\mathrm{which}\mathrm{evaluates}$$$$\mathrm{to}1.$$

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