Prove that $1 · 1! + 2 · 2! + … + n · n! = (n + 1)! − 1$ via mathematical induction?

Question

I have gotten all the way to $(k+1)!-1 +(k+1)(k+1)! = (k+2)-1$. I do not know how to proceed from here. I have seen this question in the forum before, but it did not fully explain the process. I also did not want to revive an old question (not sure if it's against the rules).

How do you go from $(k+1)! - 1 + (k+1)(k+1)!$ to $(k+1)!(1+(k+1))−1$? This has me stumped.

Answer 1

Let $a_n = 1\cdot1!+2\cdot2!+...+n\cdot n!$ and $b_n = (n+1)!-1$. Then, for $n=1$, we have $a_1 = 1 = b_1$ so the argument holds. Now, suppose inductively that $n \ge 2$ and $a_n = b_n$ for all $n$. Then, for $n+1$, we have $$a_{n+1} = 1\cdot1!+2\cdot2!+...+n\cdot n!+(n+1)\cdot (n+1)! = a_n+(n+1)\cdot (n+1)!$$ by inductive assumption and which is also $$ a_{n+1} = a_n+(n+1)\cdot (n+1)! = (n+1)!-1+(n+1)\cdot (n+1)!$$ $$ = (n+1)!(n+1+1)-1$$ $$ = (n+2)!-1$$ $$ = b_{n+1}$$ since inductive assumption claims that $a_n = b_n = (n+1)!-1$.

So we have $a_{n+1} = b_{n+1}$. Therefore by induction, argument holds for all $n$.

Answer 2

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