Proof for $\sum_{r=1}^{n}r(r!)=\sum_{r=1}^{n}[(r+1)!-r!]=(n+1)!-1$ [duplicate]

I came across the form $$\sum_{r=1}^{n}r(r!)=\sum_{r=1}^{n}[(r+1)!-r!]=(n+1)!-1$$ while solving a question in determinants. How do we get to the formula stated above?

marked as duplicate by Mike Earnest, Community♦Feb 7 at 18:50

$$\sum_{k=n_0}^{N}{(a_{k+1}-a_k)}=a_{N+1}-a_{n_0}$$ $$\underline{Derivation}$$ One can proof that by Induction. Here’s another way.

$$\sum_{k=n_0}^{N}{(a_{k+1}-a_k)}= \sum_{k=n_0}^{N}a_{k+1}-\sum_{k=n_0}^{N}{a_k}$$ $$=\sum_{k=n_{0}+1}^{N+1}a_k-\sum_{k=n_0+1}^{N}{a_k}-a_{n_0}$$ $$=a_{N +1}-a_{n_0}+\sum_{k=n_0+1}^{N}{a_k}-\sum_{k=n_0+1}^{N}{a_k}$$ $$=a_{N+1}-a_{n_0}$$