= Solution
For <Fodor's theorem>, let $\kappa$ be regular uncountable, $S\subseteq\kappa$ stationary, and $f:S\to\kappa$ regressive: $f(\alpha)<\alpha$ for every nonzero $\alpha\in S$. A <club set> is unbounded in $\kappa$ and closed under limits below $\kappa$; a <stationary set> meets every club. Removing $0$ preserves stationarity. We prove that some fibre of $f$ is stationary, not merely that one fibre is large.
We first check the club facts used in the argument. An <intersection> of fewer than $\kappa$ club <subsets> of $\kappa$ is club. Closure is immediate. For unboundedness, given $\gamma$ and clubs $(C_i)_{i<\mu}$ with $\mu<\kappa$, choose an increasing <sequence> $\gamma_n$ starting above $\gamma$, such that $\gamma_{n+1}$ is above a point of every $C_i$ beyond $\gamma_n$. Regularity bounds the <supremum> of the fewer-than-$\kappa$ points at each step. Then $\delta=\sup_{n<\omega}\gamma_n<\kappa$, and the chosen points in each $C_i$ are cofinal in $\delta$; closure puts $\delta$ in their <intersection>.
For a <sequence> $(C_\xi)_{\xi<\kappa}$ of clubs its <diagonal intersection>
$$
\Delta_{\xi<\kappa}C_\xi
=\{\delta<\kappa:\forall\xi<\delta\ (\delta\in C_\xi)\}
$$
is also club. To obtain a point above $\gamma$, recursively choose $\gamma_{n+1}>\gamma_n$ in $\bigcap_{\xi\le\gamma_n}C_\xi$, possible by the preceding result, and take the <supremum> $\delta$ of the <sequence>. For each $\xi<\delta$, every sufficiently late $\gamma_n$ belongs to $C_\xi$, so $\delta\in C_\xi$. For closure, if diagonal-intersection points are cofinal in $\delta<\kappa$, then for each $\xi<\delta$ their sufficiently late points lie in $C_\xi$, and closure again puts $\delta$ there.
Suppose every fibre $S_\xi=\{\alpha\in S:f(\alpha)=\xi\}$ were nonstationary. Choose a club $C_\xi$ disjoint from each fibre. Stationarity supplies $\alpha\in S\cap\Delta_{\xi<\kappa}C_\xi$, with $\alpha>0$. Put $\xi=f(\alpha)<\alpha$. By diagonal membership $\alpha\in C_\xi$, but by its definition $\alpha\in S_\xi$, a contradiction. Thus
$$
\boxed{\exists\xi<\kappa\quad \{\alpha\in S:f(\alpha)=\xi\}\text{ is stationary}.}
$$
This is precisely the pressing-down assertion. Regularity and uncountability were used to keep the club-building suprema below $\kappa$.
For the <Cantor normal form> theorem, every nonzero <ordinal> has a unique expression
$$
\boxed{\alpha=\omega^{\beta_0}n_0+\cdots+\omega^{\beta_{m-1}}n_{m-1},
\qquad \beta_0>\cdots>\beta_{m-1},\quad 0<n_i<\omega.}
$$
All operations are <ordinal> operations, so the order of the terms matters. The <ordinal> zero has the empty expression. We give the greedy existence proof and then prove uniqueness.
The powers $\omega^\beta$ are strictly increasing and continuous at limit exponents, and satisfy $\omega^\beta\ge\beta$. Hence the <set> of exponents with $\omega^\beta\le\alpha$ is nonempty and bounded, and its <supremum> $\beta_0$ belongs to it by continuity (or is already its last successor member). Thus there is a largest such exponent and
$$
\omega^{\beta_0}\le\alpha<\omega^{\beta_0+1}
=\omega^{\beta_0}\cdot\omega.
$$
Since the last product is the <supremum> of the finite multiples, there is a unique positive finite $n_0$ satisfying
$$
\omega^{\beta_0}n_0\le\alpha<\omega^{\beta_0}(n_0+1).
$$
The initial segment of $\alpha$ of length $\omega^{\beta_0}n_0$ leaves a tail of unique order type $\rho$, so
$$
\alpha=\omega^{\beta_0}n_0+\rho,\qquad \rho<\omega^{\beta_0}.
$$
The bound follows from strict increase of <ordinal addition> in its right argument: a tail at least $\omega^{\beta_0}$ would contradict the preceding interval. This is the relevant instance of the <ordinal division algorithm>.
If $\rho>0$, apply the same construction to it. Its leading exponent must be below $\beta_0$. Continue with the remainder. An infinite continuation would produce an infinite strictly descending <sequence> of <ordinals>, impossible by <well-foundedness>. Thus the process terminates after finitely many steps and gives the required form.
To prove uniqueness, first observe that a finite decreasing-exponent suffix with leading exponent $\gamma$ is smaller than $\omega^{\gamma+1}$. This follows by induction on its number of terms: the later tail is below $\omega^\gamma$, so adding it to a finite multiple of $\omega^\gamma$ stays below the next finite multiple and hence below $\omega^\gamma\cdot\omega$. Consequently the tail after a term with exponent $\beta$ is below $\omega^\beta$.
Any proposed normal form therefore places $\alpha$ in exactly the interval
$$
\omega^{\beta_0}n_0\le\alpha<\omega^{\beta_0}(n_0+1)
<\omega^{\beta_0+1}.
$$
Its first exponent is recoverable as the largest exponent whose power does not exceed $\alpha$, and its first coefficient is recoverable as the largest finite multiple not exceeding $\alpha$. The right remainder is unique because <ordinal addition> is strictly increasing in its right argument. Apply the same reasoning to the remainder, successively fixing every exponent and coefficient. This proves uniqueness without treating <ordinal addition> as commutative or cancelling arbitrary left summands.
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