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**clic-clac** If you consider $\displaystyle \emptyset\neq A=\{n_1,...,n_k\}\subseteq\mathbb{N},$ then the subset $\displaystyle B$ of $\displaystyle \mathbb{N}$ of elements greater than all $\displaystyle n_i, 1\leq i\leq k$ is non-empty ($\displaystyle \sum\limits_{i=1}^kn_i +1$ belongs to it)

Therefore it has a minimum, $\displaystyle m,$ and $\displaystyle m-1=\max(A).$

Indeed, if $\displaystyle m-1\notin A;$ by definition of $\displaystyle m,\ p\geq m\Rightarrow p\notin A,$ so $\displaystyle m-1>n_i,\ 1\leq i\leq n,$ i.e. $\displaystyle m-1\in B,$ contradiction.

Hence $\displaystyle m-1\in A$ and forall $\displaystyle a\in A, a\leq m-1$ by definition of $\displaystyle m,$ i.e. $\displaystyle m-1=\max(A).$

There are various ways I guess to show that, we can proceed by induction on $\displaystyle k$:

If $\displaystyle k=1,$ nothing to show $\displaystyle (n_1=\max(A)).$ Assume any subset of $\displaystyle \mathbb{N}$ with $\displaystyle k$ elements has a max, consider $\displaystyle \{n_0,...,n_k\},$ then $\displaystyle \{n_1,...,n_k\}$ has a max, let's say $\displaystyle n_j,$ two cases (because the order is total):

either $\displaystyle n_j<n_0,$ then $\displaystyle n_0=\max(\{n_0,...,n_k\})$

or $\displaystyle n_0<n_j,$ and $\displaystyle n_j=\max(\{n_0,...,n_k\})$

Actually the second proof may be better than the first: it only uses the totality of the order, which is a sufficient (and necessary) condition to be able to say that any finite non-empty subset has a maximum.