How prove this equation $\alpha=\aleph_{0}\alpha$ Question:

let $\alpha$ be a infinite cardinal , show that:
  $$\alpha=\aleph_{0}\alpha$$
  where $\aleph_{0}$ is the cardinality of the natural numbers is denoted aleph-null

this equation is from a book,and I can't prove it,and the author say this is important equation,and can't prove it.Thank you for you help
 A: First of all take well-ordered set — every non-empty subset has a least element.
Then, for every (with the exception of the greatest element) element $x$ there is the "next" element $x+1$ — the least of all which are greater than $x$.
For some elements there may be no predecessor — let us call them limit elements (or points).
And every element of a well-ordered set has a form $x=a+n$ where $a$ is a limit element, $n$ is a natural number and $+$ means the same as above.  
OK, now Zermelo's theorem, which is equivalent to an axiom of choice, every set can be well-ordered.  
Also, $|\mathbb N \times \mathbb N|=|\mathbb N|$.
And we want to prove that for infinite $A:\;A\times \mathbb N$ equinumerous to $A$.
$A$ could be well-ordered, then $A$ is equinumerous to $B\times \mathbb N$ where $B$ is the set of all limit elements.
And we have $|A\times\mathbb N|=|(B\times \mathbb N)\times\mathbb N|=|B\times(\mathbb N\times\mathbb N)|=|B\times\mathbb N|=|A|$.
A: Recall that $\aleph_\alpha$ is the cardinality of the $\alpha$-th [infinite] initial ordinal, denoted by $\omega_\alpha$.
For this proof you will need an extra fact:

If $\alpha$ is an uncountable ordinal then $|\alpha|=|\{\delta<\alpha\mid\delta\text{ is a limit ordinal}\}|$.

If $\alpha=0$ then you can prove directly that $\aleph_0\cdot\aleph_0=\aleph_0$ by using a pairing function.
Now define an equivalence relation on the ordinals below $\omega_\alpha$, $\xi\sim\zeta$ if there are only finitely many ordinals between $\xi$ and $\zeta$. I will leave you to verify that this is an equivalence relation. $\omega_\alpha$ is well-ordered, each equivalence class has a least element, and it is a limit ordinal, or zero. Otherwise its predecessor is in the same equivalence class, which is a contradiction to the minimality. And of course, different limit ordinals cannot be in the same equivalence class, because they have infinitely many ordinals between them.
So we have $\aleph_\alpha$ equivalence classes; and $\aleph_0$ points in each of the classes. Therefore $|\omega_\delta|=\aleph_\alpha\cdot\aleph_0$. Note that the axiom of choice wasn't used here, because we have a canonical ordering of each equivalence class.
Now. How to prove the fact? Usually the proof goes through defining the above equivalence relation and using the fact that $\aleph_0\cdot\aleph_\alpha=\aleph_\alpha$. So we can't really use that here. But we can use transfinite induction. It suffices to prove this for the initial ordinals, because every limit ordinal which is not an initial ordinal is either countable, or equipotent with its initial ordinal.
Suppose that for $\alpha$ it is true that for all $\beta<\alpha$, either $\beta=0$ or there are $\aleph_\beta$ limit ordinals below $\omega_\beta$. Then it holds that $\aleph_0\cdot\aleph_\beta=\aleph_\beta$ for those $\beta$'s.
By contradiction there are less than $\aleph_\alpha$ limit ordinals below $\omega_\alpha$, then there is some $\beta<\alpha$ such that $|\{\delta<\omega_\alpha\mid\delta\text{ is a limit ordinal}\}|=\aleph_\beta$. Consider the above equivalence relation defined on $\omega_\alpha$, then it only has $\aleph_\beta$ equivalence classes, and therefore $\aleph_\beta=\aleph_\beta\cdot\aleph_0=\aleph_\alpha$, but $\beta<\alpha$ and that is impossible. $\square$
So at the end we have to resort to transfinite recursion if we want to use the structure of ordinals, but it's just a scary term, it's not that scary once you understand this extremely useful technique.

Now, using the axiom of choice if $\alpha$ is an infinite cardinal then it is an $\aleph$ number. However, some people define "cardinal" as an $\aleph$ number to begin with, in which case you don't use the axiom of choice.
In the general form, where cardinal need not be an $\aleph$ number, it is consistent with the failure of the axiom of choice that there are cardinals which do not satisfy this property.
