Prove if $n \in \mathbb N$, then $\frac{1}{2!}+\cdots+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$ 
Prove if $n \in \mathbb N$, then $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{n}{\left(n+1\right)!} = 1-\frac{1}{\left(n+1\right)!}$

So I proved the base case where $n=1$ and got $\frac{1}{2}$
Then since $n=k$ implies $n=k+1$ I setup the problem like so:
$\frac{k}{(k+1)!}+\frac{(k+1)}{(k+2)!}=1-\frac{1}{(k+2)!}$
After trying to simplify it I got the following:
$\frac{k(k+2)!+(k+1)(k+1)!}{(k+1)!(k+2)!}=1-\frac{1}{(k+2)!}$
However, I'm having trouble simplifying it to match the RHS.  Hints?
 A: 
I thought it might be instructive to present an approach that is not a proof by induction.  To that end, note that we can write
$$\bbox[5px,border:2px solid #C0A000]{\frac{k}{(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!}}$$

Therefore, we have a telescoping sum such that
$$\begin{align}
\sum_{k=1}^n \frac{k}{(k+1)!}&=\sum_{k=1}^n\left(\frac{1}{k!}-\frac{1}{(k+1)!}\right)\\\\
&=1-\frac{1}{(n+1)!}
\end{align}$$
And we are done!
A: The equality you want to prove is
$$
\underbrace{\frac{1}{2!}+\dots+\frac{k}{(k+1)!}}_{*}+
\frac{k+1}{(k+2)!}=1-\frac{1}{(k+2)!}
$$
The term marked $*$ is equal, by the induction hypothesis, to
$$
1-\frac{1}{(k+1)!}
$$
and so you need to manipulate
$$
1-\frac{1}{(k+1)!}+\frac{k+1}{(k+2)!}
$$
Hint:
$$
1-\frac{1}{(k+1)!}+\frac{k+1}{(k+2)!}
=
1-\frac{k+2}{(k+2)!}+\frac{k+1}{(k+2)!}
=\dots
$$
Do the necessary steps in order to finish up at $1-\dfrac{k+1}{(k+2)!}$.
A: You have your base case.
Explicitly state your inductive hypothesis.
Suppose:
$\sum_\limits{n=1}^k\frac{n}{(n+1)!} = 1-\frac{1}{(k+1)!}$
We will show that:
$\sum_\limits{n=1}^{k+1}\frac{n}{(n+1)!} = 1-\frac{1}{(k+2)!}$
$\sum_\limits{n=1}^k\frac{n}{(n+1)!}+\frac{(k+1)}{(k+2)!}$
$1-\frac{1}{(k+1)!}+\frac{(k+1)}{(k+2)!}$(By the inductive hypothesis.)
$1-\frac{1}{(k+2)!}$
A: You can simply write it like this also this solution doesn't need induction:
$=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{n+1-1}{(n+1)!}$
$=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{1}{n!}-\frac{1}{(n+1)!}$
$=1-\frac{1}{(n+1)!}$
$$solved!$$
