Discussion
The gamma function satisfies \( \Gamma(n+1) = \)
Answer: D
Integrating \( \Gamma(n+1) = \int_{0}^{\infty} x^{n}e^{-x}dx \) by parts gives
\[ \Gamma(n+1) \;=\; \left[-x^{n}e^{-x}\right]_{0}^{\infty} + n\int_{0}^{\infty} x^{n-1}e^{-x}dx \;=\; n\,\Gamma(n). \]
With \( \Gamma(1) = 1 \) this recursion gives \( \Gamma(n+1) = n! \) for positive integers, so the function interpolates the factorials
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