The Joy of Sets, by Keith Devlin. Is it a typography error or did I not understand? I'm studying Axiomatic Set Theory with my reference book The Joy of Sets, by Keith Devlin.
I'm reporting the text on p.25:

Exercise 1.7.3. I have introduced the notation $\alpha + 1$ for the next ordinal after $\alpha$. Let us denote by $\alpha + n$ the $n$-th ordinal after $\alpha$, where $n$ is any natural number. Show that if $\alpha$ is any ordinal, either $\alpha$ is a limit ordinal or else there is a limit ordinal $\beta$ and a natural number $n$ such that $\alpha = \beta + n$

I think that Prof. Devlin meant $\beta = \alpha +n$, here. If I'm wrong, please help me understand.
 A: I think Devlin is right. He is asking you to show that if $\alpha$ is not a limit ordinal then it is a finite distance from some smaller limit ordinal.
Your reading asks you to show that if $\alpha$ is not a limit ordinal then there is a limit ordinal a finite distance on. That's clearly false - consider $2$.
A: The comments and posted answer already explain it, but here it is one more time, with a more specific example, to aid understanding. 
Say $\alpha=\omega+5$. Then $\alpha=\beta+n$ where $\beta=\omega$ is limit and $n=5$ is natural. 
In general take any ordinal $\alpha$. (In the sequel $0$ is considered both a natural number and a limit ordinal.)
If $\alpha$ is limit then $\alpha=\beta+n$ where $\beta=\alpha$ is limit and $n=0$ is natural.
If $\alpha$ is not limit then $\alpha=\gamma_1+1$ for some ordinal $\gamma_1$. If $\gamma_1$ is limit then 
$\alpha=\beta+n$ where $\beta=\gamma_1$ is limit and $n=1$ is natural.
If $\gamma_1$ is not limit then $\gamma_1=\gamma_2+1$ for some ordinal $\gamma_2$. If $\gamma_2$ is limit then $\alpha=\gamma_1+1=\gamma_2+1+1=\beta+n$ where $\beta=\gamma_2$ is limit and $n=2$ is natural.
This process cannot continue indefinitely, since there is no strictly decreasing infinite sequence of ordinals 
$\alpha>\gamma_1>\gamma_2>\gamma_3>\cdots$
Therefore there is some natural number $n$ such that $\gamma_n$ is limit and $\alpha=\gamma_n+n=\beta+n$.  
Note that, on the other hand, if $\beta = \alpha +n$ 
then the only case when $\beta$ is limit is if $n=0$ and $\alpha$ is limit (assuming $n$ stands for any natural number). Indeed, if $n\not=0$ is a natural number, then $n=m+1$ for some natural number $m$, so $\beta=(\alpha +m)+1$ is a successor.
If $\alpha=\omega+5$ and $n$ is natural, then $\beta=\alpha+n$ cannot be limit. Indeed, if $n\not=0$ then $n=m+1$ for some natural $m$ and 
$\alpha+n=(\alpha+m)+1$. If $n=0$ then $\alpha+n=(\omega+4)+1$.
Finally note that, if $\alpha=\omega+5$, then $\beta=\alpha+\omega=\omega+5+\omega=\omega+\omega$ 
so $\beta$ is limit and $\beta=\alpha+\gamma$ but at the expense of allowing $\gamma=\omega$ to be a limit ordinal. 
