I am well-acquainted with the use of mathematical induction in proofs, and I think I can usually tell when it's better to use one technique of proof over another. But up until now, the only way I've approached statements of the form "$\forall n\in\Bbb N: P(n)$" was using mathematical induction. I was taught that mathematical induction is a very useful tool, but recently I've encountered statements that seem like they could be proved much more easily without it. The problem got me thinking about this is the following:
"If $S$ is an ordered set with the least upper bound property, then any finite subset of $S$ has a supremum in $S$."
To prove it by induction, we could start with a subset $A \subseteq S$, with $A = \{x, y\}$, for some $x, y \in S$. By choosing $\alpha = max(x, y)$, we see that $\alpha$ is the supremum of $A$. Now, if we suppose that all such subsets $A_n$ have suprema given by $\alpha_n = max(\{x: x \in A_n \})$, then we can show that the proposition is still true for subsets of size $n+1$.
My only issue is that it seems like we could do this whole process directly. Without going through the induction machinery, couldn't we, right from the beginning, say something like "let $A_{n+1}\subseteq S$, such that $|A_{n+1}| = n+1$, and choose $\alpha_{n+1} = max(\{x:x \in A_{n+1} \})$," and show directly that $\alpha_{n+1}$ satisfies the properties of a supremum? In doing so, we would be able to prove the statement above without using induction, even though it's a "$\forall n\in \Bbb N$" statement.
Obviously, in case of finite subsets of ordered set, maximum and suprimum have the same meaning. Here your statement must be: "If S is an ordered set with the least upper bound property, then any finite subset A of S has a supremum in A." You've assumed $|A| = n+1$ and $ max (A) = \alpha $. This means you are assuming the result for finite sets without proving it.