TITLE: Proving that for infinite $\kappa$, $|[\kappa]^\lambda|=\kappa^\lambda$
QUESTION [9 upvotes]: Initially assume ZFC.
Let $\binom{\kappa}{\lambda}$ denotes $\left|[\kappa]^{\lambda}\right|$ where $[\kappa]^{\lambda}$ is the collection of all subsets of $\kappa$ with cardinality $\lambda$. That is, the number of subsets of $\kappa$ of size $\lambda$.
Let $\kappa$ be any infinite cardinal number. Then it is easy to see that

$\binom{\kappa}0=1$
$\binom{\kappa}n=\kappa$ for all positive natural number $n$. 
$\binom{\aleph_0}{\aleph_0}=\beth_1$
$\binom{\kappa}{\lambda}\le \kappa^\lambda$ when $1\le \lambda \le \kappa$

3 is due to $\beth_1=\sum_{\lambda=0}^{\aleph_0}\binom{\aleph_0}{\lambda}$ and others besides $\binom{\aleph_0}{\aleph_0}$ sums $\aleph_0$(since countable infinite countable infinites sums to countable infinite)
$\left({}^{\kappa}_{\lambda}\right)\le \kappa^\lambda$ is due to an injection can be made from $[\kappa]^{\lambda}$ into $\kappa^{\lambda}$(maps $\{a_i|i<\lambda\}$ to $\langle a_i\rangle_{i<\lambda}$ where $a_j<a_k$ iff $j<k$)
I hypothesize that 
$$\binom{\kappa}{\lambda}=\begin{cases}
\kappa^\lambda,&\lambda \le \kappa \\
0,& \lambda>\kappa 
\end{cases}$$
It holds when $\kappa=\aleph_0$, but hard to induct for larger cardinals without assuming GCH. So I need some helps.

REPLY [5 votes]: This becomes easier if you regard functions as sets of ordered pairs.  Then every function from $\lambda$ into $\kappa$ (the sort of thing counted by $\kappa^\lambda$) is a $\lambda$-element subset of $\kappa\times\kappa$.  Since $\kappa\times\kappa$ has $\kappa$ elements, it has $\binom\kappa\lambda$ subsets of size $\lambda$.