An alternative, simpler, proof was given by Balister, Bollobás, Morris, Sahasrabudhe, and Tiba [BBMST22], who improved the bound on the smallest modulus to $616000$.
In [Er81] offered \$1000 for a proof or disproof even just in the special case when $k=3$, which he expected 'contains the whole difficulty'. He also wrote 'I really do not see why this question is so difficult'.
The usual focus is on the regime where $k=O(1)$ is fixed (say $k=3$) and $n$ is large, although for the opposite regime Kostochka, Rödl, and Talysheva [KoRoTa99] have shown \[f(n,k)=(1+O_n(k^{-1/2^n}))k^n.\]
This was solved in the affirmative if the minimum degree is larger than some absolute constant by Liu and Montgomery [LiMo20] (therefore disproving the above stronger conjecture of Erdős and Gyárfás). Liu and Montgomery prove a much stronger result: if the average degree of $G$ is sufficiently large then there is some large integer $\ell$ such that for every even integer $m\in [(\log \ell)^8,\ell]$, $G$ contains a cycle of length $m$.
An infinite tree with minimum degree $3$ shows that the answer is trivially false for infinite graphs.
See also [3].
The first open case is $\beta=\omega^2$ (see [591]). Galvin and Larson [GaLa74] have shown that if $\beta\geq 3$ has this property then $\beta$ must be 'additively indecomposable', so that in particular $\beta=\omega^\gamma$ for some $\gamma<\omega_1$. Galvin and Larson conjecture that every $\beta\geq 3$ of this form has this property.
See also [590].
If $G$ is a random graph with $n$ vertices and each edge included independently with probability $1/2$ then is it true that almost surely \[\chi(G) - \zeta(G) \to \infty\] as $n\to \infty$?
It is known that almost surely \[\frac{n}{2\log_2n}\leq \zeta(G)\leq \chi(G)\leq (1+o(1))\frac{n}{2\log_2n}.\] (The final upper bound is due to Bollobás [Bo88]. The first inequality follows from the fact that almost surely $G$ has clique number and independence number $< 2\log_2n$.)
Heckel [He24] and, independently, Steiner [St24b] have shown that it is not the case that $\chi(G)-\zeta(G)$ is bounded with high probability, and in fact if $\chi(G)-\zeta(G) \leq f(n)$ with high probability then $f(n)\geq n^{1/2-o(1)}$ along an infinite sequence of $n$. Heckel conjectures that, with high probability, \[\chi(G)-\zeta(G) \asymp \frac{n}{(\log n)^3}.\]