Is it possible that $A\subseteq A\times B$ for some non empty sets $A,B$? I was wondering if there exist two non empty sets $A,B$ such that
$$A\subseteq A\times B.$$
I know that always exists a subset of $A\times B$ with the same cardinality of $A$, but i'm requesting here that $A$ is a subset of $A\times B$ without using any identification map. 
At first i thought that this was not possible because $A$ and $B\times A$ are two sets containing different kind of elements: the second contains pairs like $(a,b)$ with $a\in A, b\in B$; the first just single elements $a\in A$. Moreover, suppose $A\subseteq A\times B$ holds and take $a \in A$. Then $a=(a_1,b_1)$ for some $a_1 \in A, b_1\in B$. For the same reason $a_1=(a_2,b_2)$ and so $a=((a_2,b_2),b_1)$. Following this argument I got some sort of recursive infinite definition for $a$ that made me suspect something is wrong. However if I take $$A=\mathbb{N}^{\mathbb{N}} ;B=\mathbb{N}$$
is it true that $A=A\times B$ or I'm missing something? Moreover, if $A\subseteq A\times B$ can be true, are there other examples?
edit: I add another strange example: take $A=\bigcup_{i=1}^{\infty} \mathbb{N}^i $ and $B=\mathbb{N}$, then $A \times B \subset A$. This makes me think that maybe exists also an example for the other inclusion.

An element of $\Bbb N^{\Bbb N}$ is an infinite sequence of natural numbers.  Buy what do we mean by a sequence.  $n_0,n_1,\ldots n_k,\ldots$ is really nothing more than a function $f:\Bbb N\to\Bbb N$.  For example, $f(k)=n_k$.
So, $\Bbb N^{\Bbb N}=\{f:\Bbb N\to\Bbb N\}$.  So, what is an element of $A\times\Bbb N$?  It is a pair $(f,n)$. The are not elements of $A$.
