The question is:

Prove (by mathematical induction) that:

$\displaystyle \int_0^\infty{x^n}{e^{-ax}}{dx}=\frac{{n(n-1)(n-2)...1}}{a^{n+1}}}$

I've proved the equality for n=1 as my base step. For my induction step I said that S(1) is true. Suppose that S(n)=S(k). We need to prove that S(k+1) is true. So the LHS becomes:

$\displaystyle \int_0^\infty{x^{k+1}}{e^{-ax}}{dx}.$ And I used integration by parts to get the expression $\displaystyle \frac{k^4+2k^3+5k^2+2k...1}{a^{k+2}}$

However I'm not sure what the RHS expression becomes for S(k+1). I thought it might be:

$\displaystyle \frac{{k(k+1)(k-1)...1}}{a^{k+2}}$

But I don't think that's correct. Any help appreciated! Many thanks.

CP