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?

You can simply write it like this also this solution doesn't need induction:
$=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{n+1-1}{(n+1)!}$
$=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{1}{n!}-\frac{1}{(n+1)!}$
$=1-\frac{1}{(n+1)!}$
$$solved!$$
