User blog comment:BlauesWasser/I've been doing googology for a long time, but I still feel like a beginner/@comment-35470197-20180822050146

Although I am also a beginner on googology, I recommend you to enjoy explicit examples of ordinals below \(\omega^{\omega^{\omega}}\) if you do not understand fgh below it.

Could you imagine \(\omega\)? It is, roughly speaking, the ordered sequece \(0 < 1 < 2 < \cdots\).

Then could you imagine \(\omega + 1\)? Here, \(\alpha + 1\) just means adding the symbol \(\alpha\) at the end of the sequence corresponding to \(\alpha\). So since \(\omega\) corresponds to \(0 < 1 < 2 < \cdots\), \(\omega+1\) corresponds to \(0 < 1 < 2 < \cdots < \omega\). You can continue this. Say, \(\omega + 2\) corresponds to \(0 < 1 < 2 < \cdots < \omega < \omega + 1\).

Then could you imagine \(\omega + \omega\), which is the limit of \(\omega + n\)? It corresponds to \(0 < 1 < 2 < \cdots < \omega + 0 < \omega + 1 < \omega + 2\cdots\). So, the addition \(\alpha + \beta\) corresponding to the sequence given by connecting the sequences corresponding to \(\alpha\) and \(\beta\). In order to distinguish \(0,1,2,\ldots\) for \(\alpha\) and \(0,1,2,\ldots\) for \(\beta\), I replace the latter sequence by \(\alpha + 0, \alpha + 1, \alpha + 2, \ldots\) before connecting to \(\alpha\).

Now could you imagine \(\omega + \omega + \omega\)? It is \(\omega + \omega\) plus \(\omega\), and hence corresponds to the sequence given by connecting the sequence \(0 < 1 < 2 < \cdots < \omega < \omega + 1 < \omega + 2 < \cdots\) corresponding to \(\omega + \omega\) and the sequence \(\omega + \omega + 0 < \omega + \omega + 1 < \cdots\) corresponding to \(\omega\) with the modification by the string \(\omega + \omega +\) in order to distinguish the symbols)

Or it might be better to denote \(a, \omega + b, \omega + \omega + c\) by \((0,a),(1,b),(2,c)\) for short. Then \(\omega + \omega + \omega\) is described as \((0,0) < (0,1) < (0,2) < \cdots < (1,0) < (1,1) < (1,2) < \cdots < (2,0) < (2,1) < (2,2) < \cdots\). Employing such shorthands are necessary in order to describe higher ordinals.

Plus one, addition, multiplication, and powers. The way above to describe the shape can be extended in a similar way below \(\varepsilon_0 = \omega^{\omega^{\omega^{\cdot^{\cdot^{\cdot}}}}}\), if you investigate some good shorthands. Please enjoy writing down \(\omega \times \omega\), \(\omega^{\omega}\), \(\omega^{\omega^{\omega}}\), and \(\varepsilon_0\) by any shorthands. For example, if you understand \(\omega^{\omega^{\omega}}\), then Cantor's normal form is good to study the presentation of ordinals below \(\varepsilon_0\).

https://en.wikipedia.org/wiki/Ordinal_arithmetic#Cantor_normal_form