Closed Form for Factorial Sum

You're right about subtracting a term; in fact, there's a (clever) strategy called "telescoping sums" and it's particularly useful here, and you won't need induction to show it. You want terms to cancel out so that you're left with the first and last terms only.

If you want to do it yourself, then stop reading here and meditate on this idea: how can you change what's in the summation notation in order to produce a sequence of numbers such that the "middle" terms cancel out?

If you want the solution, here it is:

Let $n=(n+1)-1$, and then substitute this into your summation notation accordingly: $$S=\sum\limits_{i=1}^{n}((n+1)-1)\cdot n!$$ $$S=\sum\limits_{i=1}^{n}[(n+1)\cdot n!-n!]$$ $$S=\sum\limits_{i=1}^{n}((n+1)!-n!)$$

Working out a few terms and the very last, we immediately see: $$S=2!-1!+3!-2!+4!-3!+...+n!-(n-1)!+(n+1)!-n!$$

Which simplifies to:

$$S=(n+1)!-1$$

We can write the above relation as below:

$\sum_{k=1}^{n}k.k!=\sum_{k=1}^{n}(k+1-1)k!=\sum_{k=1}^{n}(k+1)!-\sum_{k=1}^{n}k!=\sum_{k=2}^{n+1}k!-\sum_{k=1}^{n}k!=\sum_{k=1}^{n+1}k!-\sum_{k=1}^{n}k!-1=(n+1)!-1$