Factorial Proof by Induction Question: $ \frac1{2!} + \frac2{3!} + \dots+ \frac{n}{(n+1)!} = 1 - \frac1{(n+1)!} $?

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 $n \in \mathbb{N}$.

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.