Summation involving a factorial: $1 + \sum_{j=1}^{n} j!j$ $$1 + \sum_{j=1}^{n} j!j$$
I want to find a formula for the above and then prove it by induction. The answer according to Wolfram is $(n+1)!-1$, however I have no idea how to get there. Any hints or ideas on how I should tackle this one?
 A: There is a nice interpretation of this identity in terms of uniqueness of representation in factorial base.  But to answer your comment to Theo Buehler's answer, telescoping sequences are just a thing that you should be aware of and try to look for, and the identity Theo used is actually equivalent to what Wolfram told you, so... 
A: This is a simple induction. Use $(n+1)! = (n+1) \cdot n! = n\cdot n! + n!$.
Added: I don't see any better way than to play around with the formula. Rearrange this and you have
$j \cdot j! = (j+1)! - j!$. But then you have
\begin{align*}
\sum_{j=1}^{n} j \cdot j! & = \sum_{j=1}^{n} [ (j+1)! - j!] \\
& = [2! - 1!] + [3! - 2!] + \cdots + [n! - (n-1)!] + [(n+1)! - n!]
\end{align*}
and you see that everything cancels, except the terms $- 1!$ from the first summand and $(n+1)!$ from the last summand, hence the sum must be equal to $(n+1)! - 1$.

Edit 2
Here's the argument:
We want to prove that the following statement $T(n)$ holds for all $n \in \mathbb{N}$:
\[
(n+1)! - 1 = \sum_{j=1}^{n} j \cdot j!.
\]
For $n = 1$ we have the statement $T(1)$:
\[
1 = (1+1)! - 1 = \sum_{j=1}^{1} j \cdot j! = 1\cdot 1! = 1,
\]
so this is ok. Assume that $T(n)$ holds. We want to prove $T(n+1)$:
\[
(n+2)! - 1 = \sum_{j=1}^{n+1} j \cdot j!.
\]
Start with the right hand side:
\[
\sum_{j=1}^{n+1} j \cdot j! = (n+1)\cdot (n+1)! + \sum_{j=1}^{n} j \cdot j!
\]
But the last sum is equal to $(n+1)! - 1$ by our assumption that $T(n)$ is true, so
\begin{align*}
\sum_{j=1}^{n+1} j \cdot j! & = (n+1)\cdot (n+1)! + (n+1)! - 1 \\
& = [(n+1) + 1]\cdot(n+1)! - 1 = (n+2) \cdot (n+1)! - 1\\
& = (n+2)! - 1,
\end{align*}
so $T(n+1)$ holds as well.
A: It looks like you have a closed form already. If you want to derive it rather than get it from an oracle, Gosper's algorithm will do the job.
