Existence of infinite number of infinite cardinals st: Prove the existence of infinite number of infinite cardinals 
1) $\alpha$, such that  $\alpha&lt\alpha^\aleph$ 
2) $\beta$, such that  $\beta=\beta^\aleph$
 A: The first problem is quite a bit harder than the second. 
For $1$), use König's theorem.  We still then need to show that, for example, there are infinitely many cardinals of cofinality say $\omega$, but that part is not hard.  For suppose we start at the infinite cardinal $\kappa_0$. Let $\kappa_1=2^{\kappa_0}$, $\kappa_2=2^{\kappa_1}$, and so on, and let $\kappa_\omega=\bigcup \kappa_n$. Then $\kappa_\omega$ has a right cofinality, and König's theorem applies. For the next one, start at $\kappa_\omega$.
For $2$), we can use for $\beta$ anything of shape say $\kappa^\aleph$, where $\kappa$ is an infinite cardinal.
A: Using $\aleph$ as a general cardinal might be ambiguous (there are places where it is used particularly for $2^{\aleph_0}$), the answer remains the same regardless to the intended use of $\aleph$.


*

*Recall that for every infinite $\kappa$ we have $\kappa&lt\kappa^{\operatorname{cf}(\kappa)}$. Simply show that there are infinitely many cardinals whose cofinality is $\aleph$. You can show that there exists a sequence $\alpha_0&lt\alpha_0^\aleph&lt\alpha_1&lt\ldots$ by starting the construction of $\alpha_{n+1}$ from the construction of $\alpha_n^\aleph$.

*Cardinal exponentiation has the property $\left(\kappa^\lambda\right)^\mu=\kappa^{\lambda\cdot\mu}$. Take $\alpha$ from the previous part and take $\beta=\alpha^\aleph$, now we have $\beta^\aleph=\alpha^{\aleph\cdot\aleph}=\alpha^\aleph=\beta$.
Note that for the second part you don't really need the first part, however it is easy to show that there are infinitely many $\beta$ if you use the fact that $\alpha_n^\aleph\neq\alpha_k^\aleph$ for $n\neq k$.
