I am trying to solve the following proof and could use feedback on if I did it correctly and how to reduce the the second-to-the-last step to get to the last step
Use mathematical induction to show that $$ \sum_{i=1}^{n} i \cdot i! = (n+1)! -1$$
This is my solution:
Proceed with induction. When $n=1$, the left-hand side reduces to 1 and the right-hand side becomes $2!-1 = 1$. Hence, the identity holds when $n=1$. Assume that the identity holds when $n=k$ for some integer $n \geq 1$. That is assume $$ \sum_{i=1}^{k} i \cdot i! = (k+1)!-1$$ For some integer $k \geq 1$, we want to show that the identity still holds for $n=k+1$. In other words, we want to show $$ \sum_{i=1}^{k+1} i \cdot i! = (k+2)!-1$$ From the induction hypothesis, we find $$ \sum_{i=1}^{k+1} i \cdot i! = \Bigg(\sum_{i=1}^{k} i \cdot i! \Bigg) + (k+1) \cdot (k+1)! $$ $$ \sum_{i=1}^{k+1} i \cdot i! = (k+1)! -1 + (k+1) \cdot (k+1)! $$
This is where I am stuck. I don't know how to reduce the previous step so that it looks like $$ \sum_{i=1}^{k+1} i \cdot i! = (k+2)! - 1 $$
