Proof by induction with factorials I need help with proving this: 
$$\sum_{i=1}^n \frac{i-1}{i!}=\frac{n!-1}{n!}$$
My induction hypothesis is:
$$\sum_{i=1}^{n+1} = \sum_{i=1}^n \frac{i-1}{i!}+\frac{(n+1)!-1}{(n+1)!}=\frac{(n+1)!-1}{(n+1)}$$
I tried a few things and landed here: 
$$\frac{(n+1)n!-1+n}{(n+1)n!}=\frac{(n+1)n!-1}{(n+1)n!}$$
there is one $n$ too much in my last equation and I don't know how to get rid of it.
Thanks for your help. 

"My induction hypothethesis is
$\sum_{i=1}^{n+1} = \sum_{i=1}^n \frac{i-1}{i!}+\frac{(n+1)!-1}{(n+1)!}=\frac{(n+1)!-1}{(n+1)}$"
WHY?!?!?!?!?!?!?!?
$\sum_{i=1}^{n+1}\frac {i -1}{i} = \sum_{i=1}^n \frac {i-1}{i} + \frac {n+ 1 - 1}{(n+1)!}$ and $\frac {n+1 - 1}{(n+1)!} \ne \frac{(n+1)!-1}{(n+1)!}$  And setting $n\to n+1$ will give you  $\frac {n!-1}{n!} \to \frac {(n+1)! - 1}{(n+1)!}$ and not $\frac{(n+1)! - 1}{n+1}$.
Surely you induction hypothethesis should have been
$\sum_{i=1}^{n+1}\frac {i -1}{i} = \sum_{i=1}^n \frac {i-1}{i} + \frac {n+ 1 - 1}{(n+1)!}= \frac {(n+1)! -1}{(n+1)!}$
Which is a matter of proving 
$\frac {n! - 1}{n!} + \frac {n}{(n+1)!} = \frac {(n+1)! -1}{(n+1)!}$
which should be very easy to prove:
$\frac {n! - 1}{n!} + \frac {n}{(n+1)!} =
$\frac {(n! - 1)(n+1)}{n!(n+1} + \frac {n}{(n+1)!} =$
$\frac {(n+1)! - (n+1)}{(n+1)!} + \frac n{(n+1)!} =$
$\frac {(n+1)! - n - 1 +n}{(n+1)!} = \frac {(n+1)! - 1}{(n+1)!}$.
