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. 
 A: HINT: you must prove that $$\frac{n!-1}{n!}+\frac{n+1-1}{(n+1)!}=\frac{(n+1)!-1}{(n+1)!}$$
A: Your induction hypothesis should be the formula you're trying to prove. As in:

Assume $\sum_{i=1}^n \frac{i-1}{i!}=\frac{n!-1}{n!}$

You're trying to prove the formula obtained by replacing every copy of $n$ with $n+1$ in the above formula. As in:

We wish to show that $\sum_{i=1}^{n+1} \frac{i-1}{i!}=\frac{(n+1)!-1}{(n+1)!}$

Does that help?
A: You know that $$\sum_{i=1}^n \frac{i-1}{i!}=\frac{n!-1}{n!}\ \ \ \ \  (1)$$
You want to prove if $$\sum_{i=1}^{n+1} \frac{i-1}{i!}=\frac{(n+1)!-1}{(n+1)!}$$
Then
$$\sum_{i=1}^{n} \frac{i-1}{i!}+\frac{(n+1)-1}{(n+1)!}=\frac{(n+1)!-1}{(n+1)!}\underbrace{\implies}_{(1)}\frac{n!-1}{n!}+\frac{(n+1)-1}{(n+1)!}=\frac{(n+1)!-1}{(n+1)!}$$
A: You want to change the indices on the first sum of $$\sum_{i=1}^{n+1}  \frac{n!-1}{n!}+\frac{(n+1)!-1}{(n+1)!}=\frac{(n+1)!-1}{(n+1)}$$ to $$\sum_{i=1}^{n+1}  \frac{i!-1}{i!}+\frac{(n+1)!-1}{(n+1)!}=\frac{(n+1)!-1}{(n+1)}$$
A: "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)!} =
$\frac {(n! - 1)(n+1)}{n!(n+1} + \frac {n}{(n+1)!} =$
$\frac {(n+1)! - (n+1)}{(n+1)!} + \frac n{(n+1)!} =$
$\frac {(n+1)! - n - 1 +n}{(n+1)!} = \frac {(n+1)! - 1}{(n+1)!}$.
