Using induction to prove that $\sum_{r=1}^n r\cdot r! =(n+1)! -1$

Use induction to prove that $\displaystyle\sum_{r=1}^n r\cdot r! =(n+1)! -1$

I first showed that the formula holds true for $n=1$. Then I put n as $k$ and got an expression for the sum in terms of $k$. I then found the sum till the $(k+1)$th term by adding the $(k+1)$th term to both sides of the equation and compared it to the sum expression I get by plugging $k+1$ into the sum expression given in the question - they don't match.

Solution 1:

Also a neat way to solve it without induction $$\sum_{r=1}^nr\cdot r!=\sum_{r=1}^{n}(r+1-1)\cdot r!=\sum_{r=1}^n (r+1)!-r!=(n+1)!-1$$ Since all terms cancel out except $(n+1)!$ and $-1$

Solution 2:

HINT

$$\begin{align} \\ (k+1)! -1 + (k+1)(k+1)! &= (k+1)!(1 + k+1) - 1 \\&= (k+1)!(k+2)-1\\&= ? \end{align}$$