elementary set theory-cardinality problem Let $A$ be a set. If $ \left\lvert A\right\lvert = \left\lvert A  \times \mathbb{N}\right\lvert$, then $\left\lvert  \left\{{0,1}\right\}^{A}\right\lvert = \mathbb{N}^{A}$.
Proof: It is not difficult to see that $\left\lvert  \left\{{0,1}\right\}^{A}\right\lvert \leq \mathbb{N}^{A}$. Hence, it suffices to show that $\left\lvert  \left\{{0,1}\right\}^{A}\right\lvert \geq \mathbb{N}^{A}$. But, how I show that ? Appreciate any advice.   
 A: Since $|A|=|\mathbb{N}\times A|$, we can use the implied bijection between $A$ and $\mathbb{N}\times A$ to produce a bijection between $\{0,1\}^A$ and $\{0,1\}^{\mathbb{N}\times A}$.
Thus it is sufficient to prove that $\{0,1\}^{\mathbb{N}\times A}$ has the same cardinality as $\mathbb{N}^A$.
Take a function $f$ in $\mathbb{N}^A$. Let $\phi(f)$ be the function $g: \mathbb{N}\times A\to \{0,1\}$ defined as follows:
$g(n,a)=1$ if $f(a)=n$ and $g(n,a)=0$ otherwise. 
Then the map $\phi$ is a bijection from $\mathbb{N}^A$ to $\{0,1\}^{\mathbb{N}\times A}$.
Remark: The same trick is used when one is doing the representability of recursive functions in formal arithmetic.  There, it is handy sometimes, given a function $f(x_1,x_2,\dots,x_k)$, to replace it by the predicate $P(y,x_1,x_2,\dots,x_k)$ which holds precisely if $y=f(x_1,x_2,x_k)$. And a predicate can be thought of as a function to $\{1,0\}$ (yes/no).   
A: Look for places you can replace $|A|$ with $|A| \cdot |\mathbb{N}|$.
For example, if $|A| = |A| \cdot |\mathbb{N}|$, then $2^{|A|} = 2^{|A| \cdot |\mathbb{N}|}$.
