Prove via induction $\sum_{k=2}^{n}{\frac{k-1}{k!}} = \frac{n!-1}{n!}, \forall n \in \mathbb{N}, n \ge 2$ I have to prove by induction, that 
$$\sum_{k=2}^{n}{\frac{k-1}{k!}} = \frac{n!-1}{n!}, \forall n \in \mathbb{N}, n \ge 2$$
$\begin{align}
\sum_{k=2}^{n+1}{\frac{k-1}{k!}} &= \sum_{k=2}^{n}{\frac{k-1}{k!} + \frac{(n+1)-1}{(n+1)!}} \\
&= \sum_{k=2}^{n}{\frac{k-1}{k!} + \frac{n}{(n+1)!}} \\
&= \frac{n!-1}{n!} + \frac{n}{(n+1)!} \\
&= \frac{[n!-1](n+1)!}{n!(n+1)!} + \frac{n! \cdot n}{n!(n+1)!} \\
&= \frac{[n!-1](n+1)! + n! \cdot n}{n!(n+1)!} \\
&= \frac{n!(n+1)!-(n+1)!+n!n}{n!(n+1)!} \\
\end{align}$
Question: How should I go on ? Do I have a mistake until now?

From your third line:
\begin{align}
\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}{(n+1)!} \\
&=\frac{(n+1)!-1}{(n+1)!}
\end{align}
Don't forget to check base case.
To go on from where you stop, divide numerator and denominator by $n!$.
