This statement is proved by [Vizing](https://books.google.co.kr/books?id=leL0Y5N0bFoC&pg=PA30&lpg=PA30&dq=V.%20G.%20Vizing,%20Vertex%20colorings%20with%20given%20colors%20(Russian).%20Diskret.%20Analiz.%2029%20(1976),%203%E2%80%9310.&source=bl&ots=zKFBHSs3nD&sig=ACfU3U0UAZFYC02b2_YFOK1NUSg8KP91BA&hl=ko&sa=X&ved=2ahUKEwif1YyHlODzAhVVeXAKHUviAYQQ6AF6BAgCEAM#v=onepage&q=V.%20G.%20Vizing%2C%20Vertex%20colorings%20with%20given%20colors%20(Russian).%20Diskret.%20Analiz.%2029%20(1976)%2C%203%E2%80%9310.&f=false) and [Erdos & Rubin](https://old.renyi.hu/%7Ep_erdos/1980-07.pdf) (page 30) independently.   

But I cannot find Vizing's paper (It's too old) and Erdos & Rubin's paper only says 'It is easily proved'.  

I think it is related to the proof of the following statement



> 
> $K\_{2,r}$ is $r$-choosable.
> 
> 
> 


as Erdos & Rubin's paper mentioned.  

But I have no idea with how to prove the generalized one.  

Would you help me?



$K_{k,m}$ is $k$-choosable if and only if $m<k^k$
Assume that $K\_{k,m}$ is not $k$-choosable with some lists of admissible colors. Let $A\_1$, $A\_2$, $\ldots$, $A\_k$ be sets of admissible colors in the small part (that with $k$ vertices). Choose arbitrarily colors $a\_i\in A\_i$ for all $i=1,\ldots,k$. The large part must contain a vertex with admissible colors $a\_1,\ldots,a\_k$, otherwise we may color each vertex in the large part. This yields that all $a\_i$'s are distinct, in other words all $A\_i$'s must be disjoint. Then all sequences $(a\_1,\ldots,a\_k)$ are different, and the large part must contain $k^k$ distinct vertices. On the other hand, if $m=k^k$, all $A\_i$'s are indeed disjoint, and the large part contain all such $k^k$ vertices, we can not color it.


