How to simplify the summation kk! without using induction? $$\sum_{k=1}^nk(k!)$$

I know the answer is (n+1)!-1..I can solve this question using principle of mathematical induction...but I would like to know if there is any other alternative approach
Set a permutation $\pi = \pi_1 \ldots \pi_{n+1}$ in $S_{n+1}$.

Now let $m = m(\pi)$ be the maximal index such that $\pi_1 = 1, \pi_2 = 2, \ldots, \pi_m = m$. 

The number of permutations such that $m(\pi) = m$ for $m &lt; n$ is $(n-m) (n-m)!$: now $n-m$ is the number of choices for $\pi_{m+1} \neq m+1$, and $(n-m)!$ is the number of permutations of the remaining $n-m$ numbers. 

No permutation can satisfy $m(\pi) = n$, and there is only one permutation such that $m(\pi) = n+1$. 

Since there are $(n+1)!$ permutations in $S_{n+1}$, one has

$$ (n+1)! = \sum_{m=0}^{n-1} (n-m)(n-m)! + 1 = \sum_{k=1}^n k \cdot k! + 1. $$
Add to the sum $1!+2!+3!+\ldots+n!=\sum _{k=1}^n k!$ 

You get

$(1+1!)+(2\times2! + 2!) + (3\times 3!+3!)+\ldots +(n\times n!+n!)=\\=2! +2! (2+1)+3!(3+1)+\ldots+n!(n+1)=2!+3!+4!+\ldots+(n+1)!$

Now subtract $1!+2!+3!+\ldots+n!=\sum _{k=1}^n k!$ 

You have $2!+3!+4!+\ldots+(n+1)! - 1!-2!-3!-\ldots-n!=(n+1)!-1$