Factorial related sum

Question

I noticed the following (nice!) identity:

$$ S_n:=\sum_{k=1}^n k!k = (n+1)!-1. $$

Of course it can be proved by induction, but I was curious if there is a direct way co compute it. My first try has been with Euler Gamma function:

\begin{aligned} S_n &= \sum_{k=1}^n k\int_0^\infty\mathrm{d}t\, e^{-t}t^k \\ &= \int_0^\infty\mathrm{d}t\, e^{-t}\sum_{k=1}^n k t^k\\ &= \int_0^\infty\mathrm{d}t\, e^{-t}\frac{(n t-n-1) t^{n+1}+t}{(1-t)^2}; \end{aligned}

then I have some difficulties to go on. Any ideas?

Answer 1

Note that $$k!k = k!(k+1-1)=(k+1)!-k!.$$ See how well the sum telescopes and everything cancel!