I need to use this statement in a paper. I believe I've proved it, but I would prefer to simply cite it since the proof is unrelated to the rest of the paper and distracts from the main focus of the section. I can find a lot of references to the fact that an infinite product of well-orderings is not necessarily a well-ordering, but finding this fact seems more difficult.
As a secondary matter, here's my proof:
Suppose
$$\mathcal{S}:=\bigcup_{\alpha\in A} (s_1^\alpha, s_2^\alpha,\ldots)$$
is any nonempty subset of $\mathbb{N}^{\mathbb{N}}$, with $(t_1,t_2,\ldots)$ an upper bound of $\mathcal{S}$. Note that $(t_1+1,0,\ldots)$ is also an upper bound of $\mathcal{S}$. Thus the set $\mathcal{T}_1$ of upper bounds of $\mathcal{S}$ that are zero after the first coordinate is nonempty. Similarly we define $\mathcal{T}_k$ to be the set of upper bounds of $\mathcal{S}$ that are zero after the $k$th coordinate. As $\mathcal{T}_k\supseteq \mathcal{T}_j$ whenever $k>j$, we then have $\mathcal{T}_k\not = \varnothing$ for all $k\geq 1$.
Beginning with any element of $\mathcal{T}_k$, there are only finitely many elements in $\mathcal{T}_k$ lexicographically below it. Thus each $\mathcal{T}_k$ has a minimal element $(t_1^k, t_2^k, \ldots)$.
Note that $(t_1^k,t_2^k,\ldots)$ is identical to $(t_1^{k+1},t_2^{k+1},\ldots)$ in the first $k-1$ coordinates, since $(t_1^{k+1},t_2^{k+1},\ldots, t_k^{k+1}+1,0,\ldots)$ is an element of $\mathcal{T}_k$, which can't be lower than $(t_1^k,t_2^k,\ldots)$. Similarly $(t_1^k,t_2^k,\ldots)$ is also an element of $\mathcal{T}_{k+1}$ and so can't be lower than $(t_1^{k+1},t_2^{k+1},\ldots)$. So let $r_j:=t_j^{j+1}$ be the eventual value in the $j$th coordinate.
We will now show $(r_1,r_2,\ldots)$ is the least upper bound for $\mathcal{S}$. If $(r_1,r_2,\ldots)$ was not an upper bound for $\mathcal{S}$, then there is $\alpha\in A$ with $(s_1^\alpha,s_2^\alpha,\ldots) > (r_1,r_2,\ldots)$. Suppose $m$ is the first index at which $s_m^\alpha\not = r_m$, in which case $s_m^\alpha>r_m$ by definition of lexicographic order. But then
$$(s_1^\alpha,s_2^\alpha,\ldots)> (r_1,r_2,\ldots,r_m,t_{m+1}^{m+1},0,\ldots) = (t_1^{m+1},t_2^{m+1},\ldots),$$
contradicting that the latter was an upper bound for $\mathcal{S}$. So $(r_1,r_2,\ldots)$ is an upper bound for $\mathcal{S}$.
On the other hand, if there were $(q_1,q_2,\ldots) <(r_1,r_2,\ldots)$ with $(q_1,q_2,\ldots)$ also an upper bound for $\mathcal{S}$, then let $m$ again be the first index where $(q_1,q_2,\ldots)$ and $(r_1,r_2,\ldots)$ differ, so that $q_m<r_m$. Then $(q_1,q_2,\ldots,q_m,q_{m+1}+1,0,\ldots)\in \mathcal{T}_{m+1}$, which means we would have chosen $r_m \leq q_m$. So $(r_1,r_2,\ldots)$ is the supremum of $\mathcal{S}$. As $\mathcal{S}\not = \varnothing$ was arbitrary, the lexicographic ordering on $\mathbb{N}^{\mathbb{N}}$ has the least upper bound property.
Best Answer
I don't have a reference, but here's a short proof that will fit in a footnote:
If you map the sequence $(a_0,a_1,a_2,\ldots)$ to the binary fraction $$ (0.\underbrace{11\cdots 11}_{a_0\text{ ones}}0\underbrace{11\cdots 11}_{a_1\text{ ones}}0\underbrace{11\cdots1 1}_{a_2\text{ ones}}0\ldots) _2 $$ and interpret that as a real number, then you get a bijection onto $[0,1)$ that preserves order between the lexicographic order on $\mathbb N^{\mathbb N}$ and the usual ordering of real numbers.
Since $[0,1)$ certainly has the supremum and infimum properties, so has the lexicographically ordered $\mathbb N^{\mathbb N}$.
Usually it would create problems for an argument of this kind that some real numbers have two binary fraction representations -- e.g. $\frac12$ is both $(0.1000\ldots)_2$ and $(0.0111\ldots)_2$ -- but this is neatly taken care of in this case by the fact that the numbers in the sequence are finite, so the mapping can never produce a bit sequence that end in infinite ones. It can produce a sequence that ends in infinite zeros, if all the $a_n$s from some point onward are $0$.