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Verifying the property of the Gamma function: Gamma(n) = (n-1)!

Mathematics
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\[\Gamma (\alpha)=\int\limits_{0}^{\infty}x^{\alpha-1}e^{-x}dx\] I understand how to get to the relationship \[\Gamma (\alpha)=(\alpha-1)\Gamma (\alpha-1)\] for any alpha >1 But I'm not sure how t get to :\[\Gamma (n)=(n-1)!\] for n is a positive integer
really? you have done all the hard work!
I dunno why I can't see this.. my brain is fried probably from doing the first part LOL

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Other answers:

yeah i guess so
\(n!=n(n-1)!\)
that is \(n!\) can be defined as the recursion \(1!=1\) and \(n!=n(n-1)!\)
OH i see right!! Oh thank you. Haha oh my that was fairly simple :P
Thanks :)!!
yeah way way more simple than showing \(\Gamma(\alpha)=(\alpha-1)\Gamma(\alpha -1)\)
For sure!!!
yw
like opening the jar after someone already loosened it
@satellite73 I couldn't help but wonder how do you keep your equations on the same line?

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