Fast formula for $\sum_{i=1}^{n} (i \cdot (i!))$, for arbitrary $n \in \mathbb{N}_1$

Question

I've been reading How To Prove It second edition by Daniel J. Velleman, and I've encountered an end-of-subsection exercise I can't answer. On page 286, exercise 10 of subsection 6.3 states:

"Find a formula for $\sum_{i=1}^{n} (i \cdot(i!))$ and prove that your formula is correct."

By "formula," it's talking about something similar to $\sum_{i=1}^{n} i = n(n+1)/2$, i.e. a formula with few operations for fast calculations of recursive summations with large $n$.

If I know what the formula is, I can probably figure out the proof by induction for myself, but I can't work out a hypothetical formula that works for the first five values of $n$. Here's the first few values:

$n \qquad \sum_{i=1}^{n} (i \cdot(i!))$

$1 \qquad\, 1$

$2 \qquad\, 5$

$3 \qquad\, 23$

$4 \qquad\, 119$

$5 \qquad\, 719$

Any ideas?

Answer 1

The formula you are looking for is $$(n+1)!-1$$ I noticed the following pattern: $$n=1: (1+1)!-1=(2)!-1=2-1=1$$ $$n=2: (2+1)!-1=(3)!-1=6-1=5$$ $$\vdots$$ $$n=5: (5+1)!-1=(6)!-1=720-1=719$$ I hope you can complete the proof from here.