I want to prove with mathematical induction that:
$$\sum_{i=1}^n i \cdot i! = (n+1)! - 1$$
So in the first step we define $n = 1$:
$$\sum_{i=1}^1 i \cdot i! = 1 \cdot 1! = 1 = 2! - 1$$
In the second step we define $n = n + 1$:
$$\sum_{i=1}^{n+1} i \cdot i! = \sum_{i=1}^n i \cdot i! + (n+1)(n+1)!$$ $$\implies (n+1)! - 1 + (n+1)(n+1)!$$
This can now be simplified to $(n+2)(n+1)! - 1$ but I do not see how...
We could do $(n+1)!(-1 + (n+1)) = (n+1)! \cdot n$ but still this will not be the same as above.
What did I miss?
Bodo
You are almost there.
Note that $$ (n+1)! - 1 + (n+1)(n+1)! = -1 + (n+1)!( 1 + (n+1) ) = -1+ (n+2)(n+1)! = (n+2)! - 1. $$