Is $\omega_\alpha$ a limit ordinal for any $\alpha\neq0$? In the 3th chapter of the book "Teoria de Conjunto (Curso intermedio)" by José Alfredo, Amor Montaño, Gabriela Campero Arena, Favio Ezequiel and Miranda Perea, 
having defined the set $CAR=\{k\colon k\text{ is a cardinal and } k\geq\omega\}$, I've read that "if $k\in CAR$ then $k$ is a limit ordinal" and to demonstrate this the text proceeds through the following arguments:"Be that $k\in CAR$. If $k$ were a successor ordinal, say $k=\alpha+1$, then a bijection of $k$ can be found with the previous ordinal, that is $k\sim\alpha$.  But then $\alpha<k$, which contradicts the fact that $k$ is cardinal. To give the bijection $k\sim\alpha$ is left as an exercise for the reader."
What would be the bijection between $k=\alpha+1$ and $\alpha$?
Then supposing that if $\lambda\neq0$ is a limit ordinal, then $1+\lambda=\lambda$, I found the function $\phi:k\to\alpha$ defined as 
$$\phi(\beta)=\begin{cases} 0, &\text{ if }\beta=\alpha \\ 1+\beta, &\text{ if }\beta\neq\alpha\end{cases}$$
Is the function $\phi$ a bijection between $k=\alpha+1$ and $\alpha$?
How to prove that $\phi$ is surjective onto $\alpha$?
 A: In fact, much more is true (and I think this is a situation where stating only the weaker result makes it look more mysterious):

For any infinite ordinal $\theta$ at all - limit or otherwise - we have $1+\theta=\theta$ in the sense of ordinal addition (and so a fortiori there is a bijcetion between $1+\theta$ and $\theta$). 

This is an argument by transfinite induction starting at $\omega$:


*

*The base case $\theta=\omega$ is basically Hilbert's Hotel. 

*The successor step follows by applying the induction hypothesis twice ($(\theta+1)+1\rightarrow\theta+1\rightarrow\theta$). 

*As a hint for the limit step, remember that by definition for $\lambda$ limit we have $$1+\lambda=\sup_{\beta<\lambda}(1+\beta).$$ Now think about what the induction hypothesis will say about the $(1+\beta)$s ...
So in fact your map can be more simply described as:


*

*$\varphi(\alpha)=0$,

*for $n\in\omega$, $\varphi(n)=1+n$, and

*for $\omega<\beta<\alpha$ we set $\varphi(\beta)=\beta$.
This is much easier to think about, and in particular it's easy to see that this is surjective.
