Question about $\aleph$-fixed point I am working through a proof on cardinals I found and can't reason some of the steps. 
The proposition is that there is an $\aleph$-fixed point, i.e. there is an ordinal $\alpha$ (which is necessarily a cardinal), so that $\aleph_{\alpha} = \alpha$. The proof goes as follows:
Let $\alpha_{0} = \aleph_{0}$ (or any other cardinal), $\alpha_{n + 1} = \aleph_{\alpha_{n}}$, and $\alpha = \sup \{ \alpha_{n} \mid n \in \omega \}$. Now if $\alpha = \alpha_{n}$ for some $n$, then $\alpha = \alpha_{n+1} = \aleph_{\alpha_{n}} = \aleph_{\alpha}$. Otherwise $\alpha$ is a limit ordinal and we have that $\aleph_{\alpha} = \sup \{ \aleph_{\xi} \mid \xi < \alpha\} = \sup\{ \aleph_{\alpha_{n}} \mid n \in \omega \} = \sup \{\alpha_{n + 1} \mid n \in \omega \} = \alpha$.
Now the limit case makes sense to me, but why on earth can we state that if $\alpha = \alpha_{n+1}$, then 
$$
\alpha = \alpha_{n+1} = \aleph_{\alpha_{n}} = \aleph_{\alpha}.
$$

I suppose the main issue I am having is why does $\alpha = \alpha_{n}$
  imply that $\alpha = \alpha_{n+1}$. 

After that, it is really just a matter of applying definitions.

If $\alpha=\alpha_n$, then $\alpha_{n+1}=\aleph_{\alpha_n}$ is no greater than $\alpha_n=\aleph_{\alpha_{n-1}}$ because of the $\sup$ in the definition of $\alpha$
