## View Poll Results: factorial proof

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Thanks for taking my question:

How can I prove by induction the following:

summation E (i=1 to n) of i * i! = (n+1)! -1

- thanks

2. Originally Posted by Rayk2003
Thanks for taking my question:

How can I prove by induction the following:

summation E (i=1 to n) of i * i! = (n+1)! -1

- thanks
Hi,
is it $\sum_{i=1}^n i\cdot i! = (n+1)!-1$ ?

base case: $n=1$ ... $1\cdot 1!=(1+1)!-1$ verified.

inductive step: suppose $\sum_{i=1}^n i\cdot i! = (n+1)!-1$ and let's prove it holds for $n+1$ too, i.e. we want to prove $\sum_{i=1}^{n+1} i\cdot i!=((n+1)+1)!-1$.

$\sum_{i=1}^{n+1} i\cdot i! = \sum_{i=1}^n i\cdot i! + (n+1)\cdot (n+1)!=(n+1)!-1+(n+1)\cdot (n+1)! =$
$=(n+1)!(1+(n+1))-1=(n+1)!(n+2)-1=(n+2)!-1=((n+1)+1)!-1$ inductive step verified.