This was a question from a mock paper, for my upcoming exam however my teacher unhelpfully did not post any solutions.
Prove that $\Gamma(n+1)=n!$.
Can anyone check if my proof is correct.
Thank you for reading.
-Alexis
2026-04-07 17:48:08.1775584088
The Gamma function and factorial satisfy $\Gamma(n+1) = n!$
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Note that he property $$G(n + 1) = n G(n)$$ you establish also holds for any constant multiple of $\Gamma$, including the zero function.
Since the proof you give is basically an inductive argument (it might be useful to say a little more in your solution about how this goes), it suffices to add a base case, that is, show that the identity holds for the lowest applicable value of $n$. Since $0!$ makes sense (but $(-1)!$ is not defined), one should show that $\Gamma(0 + 1) = 0!$.