[proofplan]
The three identities are exactly the limit clauses in the transfinite recursive definitions of ordinal addition, multiplication, and exponentiation. The only set-theoretic point to check is that each displayed supremum is formed from a genuine set of ordinals, not a proper class. Since the index class $\{\beta:\beta<\lambda\}$ is the ordinal $\lambda$ itself, Replacement gives the set of earlier values, and the supremum of any set of ordinals is its union.
[/proofplan]
[step:Form the sets of earlier values indexed by $\lambda$]
Fix an ordinal $\alpha$ and a nonzero limit ordinal $\lambda$. Define the sets
\begin{align*}
A_+&:=\{\alpha+\beta:\beta<\lambda\},\\
A_\cdot&:=\{\alpha\cdot\beta:\beta<\lambda\}.
\end{align*}
Since $\lambda$ is an ordinal, the collection of all $\beta$ with $\beta<\lambda$ is the set $\lambda$. The operations $\beta\mapsto \alpha+\beta$ and $\beta\mapsto \alpha\cdot\beta$ are class functions defined by transfinite recursion on ordinals, so the Replacement axiom applied to the set $\lambda$ gives that $A_+$ and $A_\cdot$ are sets. Their elements are ordinals, because ordinal addition and ordinal multiplication return ordinals.
Now let $\gamma$ be a nonzero ordinal. Define
\begin{align*}
A_{\exp}:=\{\gamma^\beta:\beta<\lambda\}.
\end{align*}
Again, Replacement applied to the set $\lambda$ and the class function $\beta\mapsto\gamma^\beta$ gives that $A_{\exp}$ is a set of ordinals.
[/step]
[step:Identify each supremum with the union of earlier values]
For any set $S$ of ordinals, define
\begin{align*}
\sup S:=\bigcup S.
\end{align*}
This union is an ordinal: if $x\in y\in\bigcup S$, then $y\in \delta$ for some $\delta\in S$, and since $\delta$ is transitive, $x\in\delta\subset\bigcup S$; also membership is well-ordered on $\bigcup S$ because it is inherited from the ordinals containing its elements.
The ordinal $\bigcup S$ is the least ordinal above every element of $S$. Indeed, if $\delta\in S$, then $\delta\subset\bigcup S$, so $\delta\leq \bigcup S$. Conversely, if $\eta$ is an ordinal such that $\delta\leq\eta$ for every $\delta\in S$, then every element of every $\delta\in S$ lies in $\eta$, hence $\bigcup S\subseteq\eta$, so $\bigcup S\leq\eta$. Therefore the displayed supremum notation is well-defined for $A_+$, $A_\cdot$, and $A_{\exp}$.
[/step]
[step:Apply the limit clause for ordinal addition]
By the transfinite recursive definition of ordinal addition with left summand $\alpha$, the value at a limit ordinal is defined by
\begin{align*}
\alpha+\lambda=\bigcup_{\beta<\lambda}(\alpha+\beta).
\end{align*}
By the preceding step, this union is precisely the supremum of the set $A_+$. Hence
\begin{align*}
\alpha+\lambda=\sup\{\alpha+\beta:\beta<\lambda\}.
\end{align*}
[/step]
[step:Apply the limit clause for ordinal multiplication]
By the transfinite recursive definition of ordinal multiplication with left factor $\alpha$, the value at a limit ordinal is defined by
\begin{align*}
\alpha\cdot\lambda=\bigcup_{\beta<\lambda}(\alpha\cdot\beta).
\end{align*}
By the supremum computation above, this union is the least ordinal greater than or equal to every $\alpha\cdot\beta$ with $\beta<\lambda$. Therefore
\begin{align*}
\alpha\cdot\lambda=\sup\{\alpha\cdot\beta:\beta<\lambda\}.
\end{align*}
[/step]
[step:Apply the limit clause for ordinal exponentiation with nonzero base]
Let $\gamma$ be a nonzero ordinal. By the transfinite recursive definition of ordinal exponentiation with base $\gamma$, the value at a nonzero limit ordinal is defined by
\begin{align*}
\gamma^\lambda=\bigcup_{\beta<\lambda}\gamma^\beta.
\end{align*}
The set of earlier values is $A_{\exp}$, and the preceding supremum argument shows that this union is exactly $\sup A_{\exp}$. Thus
\begin{align*}
\gamma^\lambda=\sup\{\gamma^\beta:\beta<\lambda\}.
\end{align*}
Combining the three limit-clause identities proves the theorem.
[/step]
