Evaluate $\sum_{k=1}^nk\cdot k!$

I discovered that the summation $\displaystyle\sum_{k=1}^n k\cdot k!$ equals $(n+1)!-1$.

But I want a proof. Could anyone give me one please? Don't worry if it uses very advanced math, I can just check it out on the internet. :)

HINT: $k(k!)=(k+1-1)(k!)=(k+1)!-k!$. Now do the summation and most of the terms will cancel.

$$\sum_{k=1}^n k\cdot k!=\sum_{k=1}^n (k+1-1)k!=\sum_{k=1}^n \left((k+1) k!- k!\right)=$$ $$=\sum_{k=1}^n ((k+1)!-k!)=\sum_{k=1}^n (k+1)!-\sum_{k=1}^n k!=$$ $$=\sum_{k=1}^{n-1} (k+1)!+(n+1)!-\sum_{k=0}^{n-1} (k+1)!=$$ $$=\sum_{k=1}^{n-1} (k+1)!+(n+1)!-(0+1)!-\sum_{k=1}^{n-1} (k+1)!=$$ $$=(n+1)!-1$$

By telescoping $$\sum_{k=1}^n k\times k!=\sum_{k=1}^n \left((k+1)\times k!- k!\right)=\sum_{k=1}^n ((k+1)!-k!)=(n+1)!-1$$

This might be a bit overkill, but I think it's still worth showing: using that $$ k!=\int_0^\infty e^{-t}t^kdt\,, $$ we have $$\begin{aligned} \sum_{k=1}^n k\cdot k!&=\int_0^\infty e^{-t}\sum_{k=1}^n kt^k dt\\ &=\int_0^\infty e^{-t}\frac{d}{dt}\sum_{k=0}^n t^k dt\\ &=\int_0^\infty te^{-t}\frac{d}{dt}\frac{1-t^{n+1}}{1-t}dt \end{aligned}$$ and integrating by parts yields $$ \sum_{k=1}^n k\cdot k! = \int_0^\infty e^{-t}(t^{n+1}-1)dt=(n+1)!-1\,. $$