So I am looking through some old exams for practice and I've run into this question where I am supposed to prove $n! =\Gamma (n+1)$ by induction.
Now I have done all the necessary work, shown that
$n\Gamma (n) =\Gamma (n+1)$
$\Gamma (1)=1 =0!$
But I'm not really sure how to structure the proof, normally when we have done induction proofs we prove the base case, $n=0$. Then assume its true for $n$, and then we show that it holds for $n +1$.
But now we already have $n+1$ in the expression, so how am I supposed to go about it? Do I prove that it holds for $n+2$? Maybe I'm missing something very obvious here but I can't seem to find it
Best Answer
Now, you assume that $n!=\Gamma(n+1)$ and then you deduce that$$(n+1)!=(n+1)\times n!=(n+1)\times\Gamma(n+1)=\Gamma(n+2).$$