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Show that $$\Gamma(1)=1$$

$$\Gamma(n+1)=\int_{0}^{\infty} e^{-t} t^{n} d t$$

$$\Gamma(1)=\Gamma(0+1)=\int_{0}^{\infty} e^{-x} x^{0} d x$$

$$=\int_{0}^{\infty} e^{-x} d x=\lim _{b \rightarrow \infty} \int_{0}^{b} \overline{e}^{x} d x$$

$$\lim _{b \rightarrow \infty} \int_{0}^{b} e^{-x} d x=\lim _{b \rightarrow \infty}\left[-e^{-x}\right]_{0}^{b}=\lim _{b \rightarrow \infty}\left[-e^{-b}-(-e^{0})\right]$$

$$=\lim _{b \rightarrow \infty}\left[-e^{-b}+1\right]=-e^{-\infty}+1=0+1=1$$

If $$n$$ is a positive integer then $$\Gamma(n+1)=n !$$

$$\Gamma(n+1)=n \Gamma(n)$$

$$=n \Gamma(n+1-1)$$

$$=n \Gamma(n-1+1)$$

$$=n \cdot(n-1) \Gamma(n-1)$$

$$=n(n-1)\Gamma(n-2+1)$$

$$=n(n-1)(n-2)\Gamma(n-2)$$

$$=n(n-1)(n-2) \cdots \cdots 3 (2) (1) \Gamma (1)=n !$$

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