Prove this by induction: $$\sum_{i=1}^n i(i!) = (n+1)!-1$$
So I wrote:
Base Case: $n=1$ so $1(1!) = 1$ and $(1+1)!-1 = 1$.
Let $n=k$ so that $$\sum_{i=1}^ki(i!)=(k+1)!-1$$
$n=k+1$ $$\sum_{i=1}^{k+1}i(i!)=((k+1)+1)!-1$$
But I'm stuck here. Also is there any easier way I can write out summation on stack exchange? A format for writing it out on one line? Or is my way preferred?
$$ \sum_1^n i\cdot i! = \sum_1^{n - 1} i\cdot i! + n\cdot n! $$
Now plug in the formula for $\sum i\cdot i!$ for $n - 1$, add the next term and see if you get the correct formula for $n$:
$$ ((n - 1) + 1)! - 1 + n\cdot n! = n! - 1 + n\cdot n! = (n + 1)n! - 1 $$
That's the inductive step. If the formula holds for $n - 1$ then it also holds for $n$.