For any integer $n$, the following fact can be proven to be true:
Given $n^n+1$ people, where each person speaks a distinct set of $n$ languages, such that any two of these people speak at least one language in common, there exists a set $T$ of $n-1$ languages such that everyone speaks at least one language in $T$.
I know how to prove that fact, and included a proof at the end, but my question is, what is the smallest number you can replace $n^n+1$ by and have that fact still be true? Letting $t_n$ be this smallest number, then $t_2=4$, for example. The fact that $t_2>3$ follows from this counterexample with three people: $$ \{\{\text{English},\text{Español}\},\{\text{English},\text{Français}\},\{\text{Español},\text{Français}\}\} $$ Furthermore, it is easy to show that for any four sets of two languages whose intersections are pairwise nonempty, there will exist a single language spoken by all. What I know so far is that $$ \binom{2n-1}{n}< t_n \le n^n+1 $$ with the lower bound realized by the collection of $n$-elements subsets of a $(2n-1)$-element set. I wonder which one of these is asymptotically correct, or if the truth is somewhere between?
The proof of $t_n\le n^n+1$ is far from obvious. You can prove the following by induction on $k$, for each $k\in \{0,1,\dots,n\}$:
Let $S$ be a set of people who each speak $n$ languages, such that every pair of people in $S$ have a language in common, and there is no set $T$ of $n-1$ languages for which everyone in $S$ speaks at least one language in $T$.
Furthermore, let $R$ be a subset of $S$ such that the intersection of all language sets of the people in $R$ has size at least $n-k$. Then $|R|\le n^k$.
In particular, $S$ itself satisfies this with $k=n$, so $|S|\le n^n$, implying any $n^n+1$ sets will have a set $T$ of $n-1$ languages which works, so $t_n\le n^n+1$.
Proof: The base case $k=0$ is obvious. For $k\ge 1$, let $R$ be a subset for which the intersection $I$ of all language sets of all people in $R$ has size at least $n-k$. We can assume $|I|\le n-1$ (else $|R|=1$), so there must be some person $p\in S$ who speaks no languages in $I$. Let $p$ speak the languages $l_1,\dots,l_n$. For each $i\in \{1,\dots,n\}$, let $R_i$ be the set of people in $R$ who speak $l_i$. Then the intersection of all the language sets for all people in $R_i$ has size at least $n-k+1$, since it includes $I\cup \{l_i\}$. By induction, we have $|R_i|\le n^{k-1}$, so $$|R|\le |R_1|+\dots+|R_n|\le n\cdot n^{k-1}=n^k.$$
You can use $n^n$ as the upper bound. This is because the proof uses $p\in S$, but in the final inductive step, it is already included in $S$, and so has already been counted. As there are no further inductive steps to make, this is not important to continue the proof.
Note that $\binom{2n-1}{n}=\frac{1}{2}\binom{2n}{n}\asymp \frac{4^n}{2\sqrt{\pi n}}$, so a reasonable asymptote might be $O(n^{n-0.5})$, but this is purely a guess!