How would you prove that via induction? $\sum_{k=1}^{n}k! \cdot k= (n+1)! -1$ Task from old exam:

Prove that $$\sum_{k=1}^{n}k! \cdot k= (n+1)! -1$$

I'd like to use induction proof and nothing else.
Here is what I did:
$n=1:$ $\sum_{k=1}^{1}1! \cdot 1= (1+1)! -1 \Leftrightarrow 1=1$
So this is true, so it will work for an $n$ already.
It worked for one $n$, now we need to show it will work for every other $n$, too.
$$\sum_{k=1}^{n+1}k! \cdot k= (n+2)! -1$$
But what can I do from this point on? There is this sum symbol on the left and I don't know how to continue with it. Any hints maybe?
I really don't want know another way of solving it. It might be easier with telescope or something else but the recommendation was induction.
 A: \begin{align}\sum_{k = 1}^{n+1} k!k &= \sum_{k = 1}^n k!k + (n+1)!(n+1) \overset{*}{=} (n+1)! - 1 + (n+1)!(n+1) \\
&= (n+1)![1 + (n+1)]-1  = \cdots\end{align}
Take it from there. Note the induction hypothesis was used at (*).
A: Note that:
$$
\begin{gathered}
  S(n) = \sum\limits_{k = 1}^n {k!k}  = \sum\limits_{k = 1}^n {k!\left( {k + 1 - 1} \right)}  =  \hfill \\
   = \sum\limits_{k = 1}^n {\left( {k + 1} \right)!}  - \sum\limits_{k = 1}^n {k!}  \hfill \\ 
\end{gathered} 
$$
then you can for sure continue
A: \begin{align*}\sum_{k=1}^{n+1} k! \cdot k &= \sum_{k=1}^n k! \cdot k + (n+1)!\cdot (n+1) \\ & = (n+1)! - 1 + (n+1)! \cdot \color{blue}{(n+1)} \\ & = (n+1)! (1 + \color{blue}{n+ 1}) - 1 \\ & = (n+2)(n+1)! - 1 \\ & = (n+2)! - 1 \end{align*}
A: Nearly always if you need to prove $\sum_{i=0}^n gingkabox(i) = floopadoop(n)$  you nearly always notice:
$\sum_{i=0}^{n+1} gingkabox(i) = floopadoop(n) + gingkabox(n+1)$
and the whole thing just becomes a matter of proving 
$floopadoop(n) + gingkabox(n+1) = floopadoop(n+1)$.
So can you prove $[(n+1)! - 1] + [(n+1)!(n+1)] = (n+2)! - 1$?
That's all you have to do.
[
$[(n+1)! - 1] + [(n+1)!(n+1)] = $
$[(n+1)! + (n+1)!(n+1)] - 1=$
$(n+1)![1 + (n+1)] -1=$
$(n+1)!(n + 2) - 1=$
$(n+2)! - 1$
]
