<s>[INST] 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?
 [/INST] 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 $1-\dfrac{k+1}{(k+2)!}$.
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