Why is this more-detailed proof more acceptable than its trivial counterpart?

Say that we're asked to give a proof of 'proof by induction'.

i.e. for some property $P$, proving that $$\forall n,P(1) \wedge [P(k) \implies P(k+1)] \implies \forall n, P(n)$$.

Now, I understand the following proof, but it seems rather superfluous:

Assume that $P(1)$ and that $P(k) \implies P(k+1).$

Assume, towards a contradiction, that there exists at least one positive integer $n$ such that $P(n)$ is false, which means the set $S=\{n:P(n) \ \text{is false}\}$ is nonempty.

By the well-ordering property, there exists a smallest element of $S$- let's call it $m$ ($\in \Bbb{Z^+}$)

Now, $m\neq 1$, since $P(1)$ is true (by assumption), so $m>1$.

Now, $m-1<m$, and $m-1 \in \mathbb{Z^+},$ so $\underbrace{m-1 \notin S}_{m \ \text{is the smallest element of }S}$, so $P(m-1) \ \text{is true}$.

But we have that $P(k) \implies P(k+1)$ by assumption, so $P(m)$ must be true (since $P(m-1)$ is true), a contradiction. Hence, the set $S=\{n:P(n) \ \text{is false}\}$ is empty. i.e. $\forall n, P(n)$. $\square$

Now, I think that this proof really labours the point. Why is it not sufficient just to say that $P(1)$ is true, and $P(k) \implies P(k+1)$, therefore, $P(2) \implies P(3) \implies P(4) \implies ... \implies P(n) \implies...$

Personally, at this level of elementary-hood, anything that includes $\ldots$ to me doesn't count as a proof, I'd rather call it a layout, plausibility argument or general description of what the actual ('social') proof would look like. (I would however accept $\ldots$ if used at a more complex level). I also reject anything related to drawings as proofs. I accept the first version as proof over the second one because it's much clearer how one would formalize it.