Factorial Proof by Induction Question: $ \frac1{2!} + \frac2{3!} + \dots+ \frac{n}{(n+1)!} = 1 - \frac1{(n+1)!} $? $\text{Use the PMI to prove the following for all natural numbers n.}$

$ \frac{1}{2!} + \frac{2}{3!} + \cdot \cdot \cdot + \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!} $

So for this question I get stuck because I cannot seem to make the left side and right side be equivalent.  
$\color{Green}{Proof} :$ 
$\mathbf (I) \; Basis \; Step :$  Show $p(1).$ For $n=1$
$\frac{1}{(1+1)!} = 1- \frac{1}{(1+1)!} \Rightarrow \frac12 = \frac12$
$\mathbf (II) \; Inductive \; Step : $ 
Assume $p(k)$ for a PAC $k \in ℕ ,\; \; \;\; n=k$
$ \frac{1}{2!} + \frac{2}{3!} +\cdot\cdot\cdot + \frac{k}{(k+1)!} = 1-\frac{1}{(k+1)!} \; \;  \text{Induction Hypothesis}$
Left Hand Side : We need to show : $p(k+1)$ i.e. $n = k+1$
$\frac{k}{(k+1)!} + \frac{k+1}{(k+1+1)!} \Rightarrow \ $
$\frac{(k+1)}{(k+1+1)!} +1 - \frac{1}{(k+1)!} $
Right Hand Side: $ n =k+1$ 
$\left(  1- \frac{1}{(k+1+1)!}\right) = 1- \frac{1}{(k+2)!}$
This is where I have made my mistake because I think I have made an arithmetic mistake on the left side because I cannot get them to be verifiable. Any hints on how to correct the  error would be appreciated. Have a good one.
 A: By the inductive hypothesis, we know that \begin{align*} \frac{1}{2!} + \cdots + \frac{k}{(k+1)!} + \frac{k+1}{(k+2)!} &= 1 - \frac{1}{(k+1)!} + \frac{k+1}{(k+2)!} \\ &= 1 - \frac{k+2}{(k+2)!} + \frac{k+1}{(k+2)!} \\ &= 1 - \frac{1}{(k+2)!}.\end{align*} Hence, if the statement is true for $n=k$, then it true for $n = k+1$, so by induction, it is true for all $n \in \mathbb{N}$.
A: What you need to show in the induction step is that
$$\frac1{2!}+\frac2{3!}+\ldots+\frac{k}{(k+1)!}+\frac{k+1}{(k+2)!}=1-\frac1{(k+2)!}\;.$$
Since your induction hypothesis is that the sum of the first $k$ terms on the left is $1-\frac1{(k+1)!}$, this amounts to showing that
$$1-\frac1{(k+1)!}+\frac{k+1}{(k+2)!}=1-\frac1{(k+2)!}\;.$$
If you multiply $\frac1{(k+1)!}$ by $1$ in the carefully chosen form $\frac{k+2}{k+2}$, you should have little difficulty showing this.
A: Although this post does not address the specific issue posed in the OP, I thought it might be instructive to present an alternative way forward.  To that end, we note that
$$\begin{align}
\frac{k}{(k+1)!}&=\frac{(k+1)-1}{(k+1)!}\\\\
&=\frac{1}{k!}-\frac{1}{(k+1)!}
\end{align}$$
Now, the sum becomes a telescoping sum and we have immediately
$$\sum_{k=1}^n\frac{k}{(k+1)!}=1-\frac{1}{(n+1)!}$$
as was to be shown!
