How do I prove this identity for summations?

Question

How would I go about proving this identity? I really have no idea where to even start, so any help would be greatly appreciated!

$$\sum_{i=0}^n \left( i + 1 \right) \left( \left( i + 1 \right) ! \right) = \left( n + 2 \right) ! - 1$$

Answer 1

Hint: $(i+1)(i+1)!=(i+2)!-(i+1)!$

Now the sum becomes $$2!-1!+3!-2!+...+(n+1)!-n!+(n+2)!-(n-1)!=(n-2)!-1!=(n-2)!-1$$

Answer 2

Try using induction by letting $\mathrm P(n)$ be the proposition that $$\sum_{i=0}^n = (i+1)(i+1)! = (n+2)!$$

Your base case $\mathrm P(0)$ is $$(1)(1)!=2!-1$$

Now you just have to prove that $\mathrm P(k)\Rightarrow\mathrm P(k+1)$. Can you go from there?