<s>[INST] Proof by mathematical induction proof verification If $n\in\Bbb N$ then $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$.
The base case where $n=1$ is clearly true: $\frac{1}{2!}=\frac{1}{2}$ and $1-\frac{1}{(1+1)!}=\frac{1}{2}$.
Now to show that $S_k\Rightarrow S_{k+1}$, we assume that  $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{k}{(k+1)!}=1-\frac{1}{(k+1)!}$ and observe the following equalitites:
$\begin{align*}\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{k+1}{((k+1)+1)!}&=\\
(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{k}{(k+1)!})+\Big(\frac{(k+1)}{((k+1)+1)!}\Big)&=\\
1-\frac{1}{(k+1)!}+\frac{k+1}{((k+1)+1)!}&=\frac{(k+2)(k+1)!-(k+2)+(k+1)}{(k+2)(k+1)!}\\
&=\frac{(k+2)(k+1)!-1}{(k+2)(k+1)!}\\
&=1-\frac{1}{(k+2)!}\\
&=1-\frac{1}{((k+1)+1)!}\end{align*}$
Therefore we conclude that if $n\in\Bbb N$ then $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$.
I know this is a relatively straightforward example but I'm very new to the concept of induction and I'm wondering if I'm on the right track with this proof. It seems correct to me but I'm worried I may have overlooked something. Any advice is helpful, thanks!
 [/INST] A couple of your steps could possibly be arranged in a more readable way. However, I do believe your proof is core.
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